Python 实用宝典

Question 1

When you have a model field with a choices option you tend to have some magic values associated with human readable names. Is there in Django a convenient way to set these fields by the human readable name instead of the value?

Consider this model:

class Thing(models.Model):
  PRIORITIES = (
    (0, 'Low'),
    (1, 'Normal'),
    (2, 'High'),
  )

  priority = models.IntegerField(default=0, choices=PRIORITIES)

At some point we have a Thing instance and we want to set its priority. Obviously you could do,

thing.priority = 1

But that forces you to memorize the Value-Name mapping of PRIORITIES. This doesn’t work:

thing.priority = 'Normal' # Throws ValueError on .save()

Currently I have this silly workaround:

thing.priority = dict((key,value) for (value,key) in Thing.PRIORITIES)['Normal']

but that’s clunky. Given how common this scenario could be I was wondering if anyone had a better solution. Is there some field method for setting fields by choice name which I totally overlooked?

Question 2

Do as seen here. Then you can use a word that represents the proper integer.

Like so:

LOW = 0
NORMAL = 1
HIGH = 2
STATUS_CHOICES = (
    (LOW, 'Low'),
    (NORMAL, 'Normal'),
    (HIGH, 'High'),
)

Then they are still integers in the DB.

Usage would be thing.priority = Thing.NORMAL

Question 3

I’d probably set up the reverse-lookup dict once and for all, but if I hadn’t I’d just use:

thing.priority = next(value for value, name in Thing.PRIORITIES
                      if name=='Normal')

which seems simpler than building the dict on the fly just to toss it away again;-).

Question 4

Here’s a field type I wrote a few minutes ago that I think does what you want. Its constructor requires an argument ‘choices’, which may be either a tuple of 2-tuples in the same format as the choices option to IntegerField, or instead a simple list of names (ie ChoiceField((‘Low’, ‘Normal’, ‘High’), default=’Low’) ). The class takes care of the mapping from string to int for you, you never see the int.

  class ChoiceField(models.IntegerField):
    def __init__(self, choices, **kwargs):
        if not hasattr(choices[0],'__iter__'):
            choices = zip(range(len(choices)), choices)

        self.val2choice = dict(choices)
        self.choice2val = dict((v,k) for k,v in choices)

        kwargs['choices'] = choices
        super(models.IntegerField, self).__init__(**kwargs)

    def to_python(self, value):
        return self.val2choice[value]

    def get_db_prep_value(self, choice):
        return self.choice2val[choice]

Question 5

I appreciate the constant defining way but I believe Enum type is far best for this task. They can represent integer and a string for an item in the same time, while keeping your code more readable.

Enums were introduced to Python in version 3.4. If you are using any lower (such as v2.x) you can still have it by installing the backported package: pip install enum34.

# myapp/fields.py
from enum import Enum    


class ChoiceEnum(Enum):

    @classmethod
    def choices(cls):
        choices = list()

        # Loop thru defined enums
        for item in cls:
            choices.append((item.value, item.name))

        # return as tuple
        return tuple(choices)

    def __str__(self):
        return self.name

    def __int__(self):
        return self.value


class Language(ChoiceEnum):
    Python = 1
    Ruby = 2
    Java = 3
    PHP = 4
    Cpp = 5

# Uh oh
Language.Cpp._name_ = 'C++'

This is pretty much all. You can inherit the ChoiceEnum to create your own definitions and use them in a model definition like:

from django.db import models
from myapp.fields import Language

class MyModel(models.Model):
    language = models.IntegerField(choices=Language.choices(), default=int(Language.Python))
    # ...

Querying is icing on the cake as you may guess:

MyModel.objects.filter(language=int(Language.Ruby))
# or if you don't prefer `__int__` method..
MyModel.objects.filter(language=Language.Ruby.value)

Representing them in string is also made easy:

# Get the enum item
lang = Language(some_instance.language)

print(str(lang))
# or if you don't prefer `__str__` method..
print(lang.name)

# Same as get_FOO_display
lang.name == some_instance.get_language_display()

Question 6

class Sequence(object):
    def __init__(self, func, *opts):
        keys = func(len(opts))
        self.attrs = dict(zip([t[0] for t in opts], keys))
        self.choices = zip(keys, [t[1] for t in opts])
        self.labels = dict(self.choices)
    def __getattr__(self, a):
        return self.attrs[a]
    def __getitem__(self, k):
        return self.labels[k]
    def __len__(self):
        return len(self.choices)
    def __iter__(self):
        return iter(self.choices)
    def __deepcopy__(self, memo):
        return self

class Enum(Sequence):
    def __init__(self, *opts):
        return super(Enum, self).__init__(range, *opts)

class Flags(Sequence):
    def __init__(self, *opts):
        return super(Flags, self).__init__(lambda l: [1<<i for i in xrange(l)], *opts)

Use it like this:

Priorities = Enum(
    ('LOW', 'Low'),
    ('NORMAL', 'Normal'),
    ('HIGH', 'High')
)

priority = models.IntegerField(default=Priorities.LOW, choices=Priorities)

Question 7

As of Django 3.0, you can use:

class ThingPriority(models.IntegerChoices):
    LOW = 0, 'Low'
    NORMAL = 1, 'Normal'
    HIGH = 2, 'High'


class Thing(models.Model):
    priority = models.IntegerField(default=ThingPriority.LOW, choices=ThingPriority.choices)

# then in your code
thing = get_my_thing()
thing.priority = ThingPriority.HIGH

Question 8

Simply replace your numbers with the human readable values you would like. As such:

PRIORITIES = (
('LOW', 'Low'),
('NORMAL', 'Normal'),
('HIGH', 'High'),
)

This makes it human readable, however, you’d have to define your own ordering.

Question 9

My answer is very late and might seem obvious to nowadays-Django experts, but to whoever lands here, i recently discovered a very elegant solution brought by django-model-utils: https://django-model-utils.readthedocs.io/en/latest/utilities.html#choices

This package allows you to define Choices with three-tuples where:

The first item is the database value
The second item is a code-readable value
The third item is a human-readable value

So here’s what you can do:

from model_utils import Choices

class Thing(models.Model):
    PRIORITIES = Choices(
        (0, 'low', 'Low'),
        (1, 'normal', 'Normal'),
        (2, 'high', 'High'),
      )

    priority = models.IntegerField(default=PRIORITIES.normal, choices=PRIORITIES)

thing.priority = getattr(Thing.PRIORITIES.Normal)

This way:

You can use your human-readable value to actually choose the value of your field (in my case, it’s useful because i’m scraping wild content and storing it in a normalized way)
A clean value is stored in your database
You have nothing non-DRY to do ;)

Enjoy :)

Question 10

Originally I used a modified version of @Allan’s answer:

from enum import IntEnum, EnumMeta

class IntegerChoiceField(models.IntegerField):
    def __init__(self, choices, **kwargs):
        if hasattr(choices, '__iter__') and isinstance(choices, EnumMeta):
            choices = list(zip(range(1, len(choices) + 1), [member.name for member in list(choices)]))

        kwargs['choices'] = choices
        super(models.IntegerField, self).__init__(**kwargs)

    def to_python(self, value):
        return self.choices(value)

    def get_db_prep_value(self, choice):
        return self.choices[choice]

models.IntegerChoiceField = IntegerChoiceField

GEAR = IntEnum('GEAR', 'HEAD BODY FEET HANDS SHIELD NECK UNKNOWN')

class Gear(Item, models.Model):
    # Safe to assume last element is largest value member of an enum?
    #type = models.IntegerChoiceField(GEAR, default=list(GEAR)[-1].name)
    largest_member = GEAR(max([member.value for member in list(GEAR)]))
    type = models.IntegerChoiceField(GEAR, default=largest_member)

    def __init__(self, *args, **kwargs):
        super(Gear, self).__init__(*args, **kwargs)

        for member in GEAR:
            setattr(self, member.name, member.value)

print(Gear().HEAD, (Gear().HEAD == GEAR.HEAD.value))

Simplified with the django-enumfields package package which I now use:

from enumfields import EnumIntegerField, IntEnum

GEAR = IntEnum('GEAR', 'HEAD BODY FEET HANDS SHIELD NECK UNKNOWN')

class Gear(Item, models.Model):
    # Safe to assume last element is largest value member of an enum?
    type = EnumIntegerField(GEAR, default=list(GEAR)[-1])
    #largest_member = GEAR(max([member.value for member in list(GEAR)]))
    #type = EnumIntegerField(GEAR, default=largest_member)

    def __init__(self, *args, **kwargs):
        super(Gear, self).__init__(*args, **kwargs)

        for member in GEAR:
            setattr(self, member.name, member.value)

Question 11

Model’s choices option accepts a sequence consisting itself of iterables of exactly two items (e.g. [(A, B), (A, B) …]) to use as choices for this field.

In addition, Django provides enumeration types that you can subclass to define choices in a concise way:

class ThingPriority(models.IntegerChoices):
    LOW = 0, _('Low')
    NORMAL = 1, _('Normal')
    HIGH = 2, _('High')

class Thing(models.Model):
    priority = models.IntegerField(default=ThingPriority.NORMAL, choices=ThingPriority.choices)

Django supports adding an extra string value to the end of this tuple to be used as the human-readable name, or label. The label can be a lazy translatable string.

   # in your code 
   thing = get_thing() # instance of Thing
   thing.priority = ThingPriority.LOW

Note: you can use that using ThingPriority.HIGH, ThingPriority.['HIGH'], or ThingPriority(0) to access or lookup enum members.

Question 12

I am trying to develop a website using Django framework and launched using DigitalOcean.com and deployed the necessary files into django-project.

I had to include static files into Django-project and After collecting static files, I tried to refresh my ip

I am including the tutorials which I have used to create the website. https://www.pythonprogramming.net/django-web-server-publish-tutorial/

I am getting the following error :

DisallowedHost at / Invalid HTTP_HOST header: ‘198.211.99.20’. You may need to add u’198.211.99.20′ to ALLOWED_HOSTS.

Can somebody help me to fix this ? This is my first website using Django framework.

Question 13

The error log is straightforward. As it suggested,You need to add 198.211.99.20 to your ALLOWED_HOSTS setting.

In your project settings.py file,set ALLOWED_HOSTS like this :

ALLOWED_HOSTS = ['198.211.99.20', 'localhost', '127.0.0.1']

For further reading read from here.

Question 14

settings.py

ALLOWED_HOSTS = ['*']

Question 15

I would like to provide the same content inside 2 different base files.

So I’m trying to do this:

page1.html:

{% extends "base1.html" %}
{% include "commondata.html" %}

page2.html:

{% extends "base2.html" %} 
{% include "commondata.html" %}

The problem is that I can’t seem to use both extends and include. Is there some way to do that? And if not, how can I accomplish the above?

commondata.html overrides a block that is specified in both base1.html and base2.html

The purpose of this is to provide the same page in both pdf and html format, where the formatting is slightly different. The above question though simplifies what I’m trying to do so if I can get an answer to that it will solve my problem.

Question 16

When you use the extends template tag, you’re saying that the current template extends another — that it is a child template, dependent on a parent template. Django will look at your child template and use its content to populate the parent.

Everything that you want to use in a child template should be within blocks, which Django uses to populate the parent. If you want use an include statement in that child template, you have to put it within a block, for Django to make sense of it. Otherwise it just doesn’t make sense and Django doesn’t know what to do with it.

The Django documentation has a few really good examples of using blocks to replace blocks in the parent template.

https://docs.djangoproject.com/en/dev/ref/templates/language/#template-inheritance

Question 17

From Django docs:

The include tag should be considered as an implementation of “render this subtemplate and include the HTML”, not as “parse this subtemplate and include its contents as if it were part of the parent”. This means that there is no shared state between included templates — each include is a completely independent rendering process.

So Django doesn’t grab any blocks from your commondata.html and it doesn’t know what to do with rendered html outside blocks.

Question 18

This should do the trick for you: put include tag inside of a block section.

page1.html:

{% extends "base1.html" %}

{% block foo %}
   {% include "commondata.html" %}
{% endblock %}

page2.html:

{% extends "base2.html" %}

{% block bar %}
   {% include "commondata.html" %}
{% endblock %}

Question 19

More info about why it wasn’t working for me in case it helps future people:

The reason why it wasn’t working is that {% include %} in django doesn’t like special characters like fancy apostrophe. The template data I was trying to include was pasted from word. I had to manually remove all of these special characters and then it included successfully.

Question 20

You can’t pull in blocks from an included file into a child template to override the parent template’s blocks. However, you can specify a parent in a variable and have the base template specified in the context.

From the documentation:

{% extends variable %} uses the value of variable. If the variable evaluates to a string, Django will use that string as the name of the parent template. If the variable evaluates to a Template object, Django will use that object as the parent template.

Instead of separate “page1.html” and “page2.html”, put {% extends base_template %} at the top of “commondata.html”. And then in your view, define base_template to be either “base1.html” or “base2.html”.

Question 21

Added for reference to future people who find this via google: You might want to look at the {% overextend %} tag provided by the mezzanine library for cases like this.

Question 22

Edit 10th Dec 2015: As pointed out in the comments, ssi is deprecated since version 1.8. According to the documentation:

This tag has been deprecated and will be removed in Django 1.10. Use the include tag instead.

In my opinion, the right (best) answer to this question is the one from podshumok, as it explains why the behaviour of include when used along with inheritance.

However, I was somewhat surprised that nobody mentioned the ssi tag provided by the Django templating system, which is specifically designed for inline including an external piece of text. Here, inline means the external text will not be interpreted, parsed or interpolated, but simply “copied” inside the calling template.

Please, refer to the documentation for further details (be sure to check your appropriate version of Django in the selector at the lower right part of the page).

https://docs.djangoproject.com/en/dev/ref/templates/builtins/#ssi

From the documentation:

ssi
Outputs the contents of a given file into the page.
Like a simple include tag, {% ssi %} includes the contents of another file
– which must be specified using an absolute path – in the current page

Beware also of the security implications of this technique and also of the required ALLOWED_INCLUDE_ROOTS define, which must be added to your settings files.

Question 23

I am building a project in Django Rest Framework where users can login to view their wine cellar. My ModelViewSets were working just fine and all of a sudden I get this frustrating error:

Could not resolve URL for hyperlinked relationship using view name “user-detail”. You may have failed to include the related model in your API, or incorrectly configured the lookup_field attribute on this field.

The traceback shows:

    [12/Dec/2013 18:35:29] "GET /bottles/ HTTP/1.1" 500 76677
Internal Server Error: /bottles/
Traceback (most recent call last):
  File "/Users/bpipat/.virtualenvs/usertest2/lib/python2.7/site-packages/django/core/handlers/base.py", line 114, in get_response
    response = wrapped_callback(request, *callback_args, **callback_kwargs)
  File "/Users/bpipat/.virtualenvs/usertest2/lib/python2.7/site-packages/rest_framework/viewsets.py", line 78, in view
    return self.dispatch(request, *args, **kwargs)
  File "/Users/bpipat/.virtualenvs/usertest2/lib/python2.7/site-packages/django/views/decorators/csrf.py", line 57, in wrapped_view
    return view_func(*args, **kwargs)
  File "/Users/bpipat/.virtualenvs/usertest2/lib/python2.7/site-packages/rest_framework/views.py", line 399, in dispatch
    response = self.handle_exception(exc)
  File "/Users/bpipat/.virtualenvs/usertest2/lib/python2.7/site-packages/rest_framework/views.py", line 396, in dispatch
    response = handler(request, *args, **kwargs)
  File "/Users/bpipat/.virtualenvs/usertest2/lib/python2.7/site-packages/rest_framework/mixins.py", line 96, in list
    return Response(serializer.data)
  File "/Users/bpipat/.virtualenvs/usertest2/lib/python2.7/site-packages/rest_framework/serializers.py", line 535, in data
    self._data = [self.to_native(item) for item in obj]
  File "/Users/bpipat/.virtualenvs/usertest2/lib/python2.7/site-packages/rest_framework/serializers.py", line 325, in to_native
    value = field.field_to_native(obj, field_name)
  File "/Users/bpipat/.virtualenvs/usertest2/lib/python2.7/site-packages/rest_framework/relations.py", line 153, in field_to_native
    return self.to_native(value)
  File "/Users/bpipat/.virtualenvs/usertest2/lib/python2.7/site-packages/rest_framework/relations.py", line 452, in to_native
    raise Exception(msg % view_name)
Exception: Could not resolve URL for hyperlinked relationship using view 
name "user-detail". You may have failed to include the related model in 
your API, or incorrectly configured the `lookup_field` attribute on this 
field.

I have a custom email user model and the bottle model in models.py is:

class Bottle(models.Model):    
      wine = models.ForeignKey(Wine, null=False)
      user = models.ForeignKey(User, null=False, related_name='bottles')

My serializers:

class BottleSerializer(serializers.HyperlinkedModelSerializer):

    class Meta:
        model = Bottle
        fields = ('url', 'wine', 'user')

class UserSerializer(serializers.ModelSerializer):

    class Meta:
        model = User
        fields = ('email', 'first_name', 'last_name', 'password', 'is_superuser')

My views:

class BottleViewSet(viewsets.ModelViewSet):
    """
    API endpoint that allows bottles to be viewed or edited.
    """
    queryset = Bottle.objects.all()
    serializer_class = BottleSerializer

class UserViewSet(ListCreateAPIView):
    """
    API endpoint that allows users to be viewed or edited.
    """
    queryset = User.objects.all()
    serializer_class = UserSerializer

and finally the url:

router = routers.DefaultRouter()
router.register(r'bottles', views.BottleViewSet, base_name='bottles')

urlpatterns = patterns('',
    url(r'^', include(router.urls)),
    # ...

I don’t have a user detail view and I don’t see where this issue could come from. Any ideas?

Thanks

Question 24

Because it’s a HyperlinkedModelSerializer your serializer is trying to resolve the URL for the related User on your Bottle.
As you don’t have the user detail view it can’t do this. Hence the exception.

Would not just registering the UserViewSet with the router solve your issue?
You could define the user field on your BottleSerializer to explicitly use the UserSerializer rather than trying to resolve the URL. See the serializer docs on dealing with nested objects for that.

Question 25

I came across this error too and solved it as follows:

The reason is I forgot giving “**-detail” (view_name, e.g.: user-detail) a namespace. So, Django Rest Framework could not find that view.

There is one app in my project, suppose that my project name is myproject, and the app name is myapp.

There is two urls.py file, one is myproject/urls.py and the other is myapp/urls.py. I give the app a namespace in myproject/urls.py, just like:

url(r'', include(myapp.urls, namespace="myapp")),

I registered the rest framework routers in myapp/urls.py, and then got this error.

My solution was to provide url with namespace explicitly:

class UserSerializer(serializers.HyperlinkedModelSerializer):
    url = serializers.HyperlinkedIdentityField(view_name="myapp:user-detail")

    class Meta:
        model = User
        fields = ('url', 'username')

And it solved my problem.

Question 26

Maybe someone can have a look at this : http://www.django-rest-framework.org/api-guide/routers/

If using namespacing with hyperlinked serializers you’ll also need to ensure that any view_name parameters on the serializers correctly reflect the namespace. For example:

urlpatterns = [
    url(r'^forgot-password/$', ForgotPasswordFormView.as_view()),
    url(r'^api/', include(router.urls, namespace='api')),
]

you’d need to include a parameter such as view_name='api:user-detail' for serializer fields hyperlinked to the user detail view.

class UserSerializer(serializers.HyperlinkedModelSerializer):
    url = serializers.HyperlinkedIdentityField(view_name="api:user-detail")

class Meta:
    model = User
    fields = ('url', 'username')

Question 27

Another nasty mistake that causes this error is having the base_name unnecessarily defined in your urls.py. For example:

router.register(r'{pathname}', views.{ViewName}ViewSet, base_name='pathname')

This will cause the error noted above. Get that base_name outta there and get back to a working API. The code below would fix the error. Hooray!

router.register(r'{pathname}', views.{ViewName}ViewSet)

However, you probably didn’t just arbitrarily add the base_name, you might have done it because you defined a custom def get_queryset() for the View and so Django mandates that you add the base_name. In this case you’ll need to explicitly define the ‘url’ as a HyperlinkedIdentityField for the serializer in question. Notice we are defining this HyperlinkedIdentityField ON THE SERIALIZER of the view that is throwing the error. If my error were “Could not resolve URL for hyperlinked relationship using view name “study-detail”. You may have failed to include the related model in your API, or incorrectly configured the lookup_field attribute on this field.” I could fix this with the following code.

My ModelViewSet (the custom get_queryset is why I had to add the base_name to the router.register() in the first place):

class StudyViewSet(viewsets.ModelViewSet):
    serializer_class = StudySerializer

    '''custom get_queryset'''
    def get_queryset(self):
        queryset = Study.objects.all()
        return queryset

My router registration for this ModelViewSet in urls.py:

router.register(r'studies', views.StudyViewSet, base_name='studies')

AND HERE’S WHERE THE MONEY IS! Then I could solve it like so:

class StudySerializer(serializers.HyperlinkedModelSerializer):
    url = serializers.HyperlinkedIdentityField(view_name="studies-detail")
    class Meta:
        model = Study
        fields = ('url', 'name', 'active', 'created',
              'time_zone', 'user', 'surveys')

Yep. You have to explicitly define this HyperlinkedIdentityField on itself for it to work. And you need to make sure that the view_name defined on the HyperlinkedIdentityField is the same as you defined on the base_name in urls.py with a ‘-detail’ added after it.

Question 28

This code should work, too.

class BottleSerializer(serializers.HyperlinkedModelSerializer):

  user = UserSerializer()

  class Meta:
    model = Bottle
    fields = ('url', 'wine', 'user')

Question 29

I ran into this error after adding namespace to my url

 url('api/v2/', include('api.urls', namespace='v2')),

and adding app_name to my urls.py

I resolved this by specifying NamespaceVersioning for my rest framework api in settings.py of my project

REST_FRAMEWORK = {
    'DEFAULT_VERSIONING_CLASS':'rest_framework.versioning.NamespaceVersioning'}

Question 30

Today, I got the same error and below changes rescue me.

Change

class BottleSerializer(serializers.HyperlinkedModelSerializer):

to:

 class BottleSerializer(serializers.ModelSerializer):

Question 31

Same Error, but different reason:

I define a custom user model, nothing new field:

from django.contrib.auth.models import (AbstractUser)
class CustomUser(AbstractUser):
    """
    custom user, reference below example
    https://github.com/jonathanchu/django-custom-user-example/blob/master/customuser/accounts/models.py

    # original User class has all I need
    # Just add __str__, not rewrite other field
    - id
    - username
    - password
    - email
    - is_active
    - date_joined
    - method, email_user
    """

    def __str__(self):
        return self.username

This is my view function:

from rest_framework import permissions
from rest_framework import viewsets
from .models import (CustomUser)
class UserViewSet(viewsets.ModelViewSet):
    permission_classes = (permissions.AllowAny,)
    serializer_class = UserSerializer

    def get_queryset(self):
        queryset = CustomUser.objects.filter(id=self.request.user.id)
        if self.request.user.is_superuser:
            queryset = CustomUser.objects.all()
        return queryset

Since I didn’t give queryset directly in UserViewSet, I have to set base_name when I register this viewset. This is where my error message caused by urls.py file:

from myapp.views import (UserViewSet)
from rest_framework.routers import DefaultRouter
router = DefaultRouter()
router.register(r'users', UserViewSet, base_name='customuser')  # <--base_name needs to be 'customuser' instead of 'user'

You need a base_name same as your model name – customuser.

Question 32

If you’re extending the GenericViewSet and ListModelMixin classes, and have the same error when adding the url field in the list view, it’s because you’re not defining the detail view. Be sure you’re extending the RetrieveModelMixin mixin:

class UserViewSet (mixins.ListModelMixin,
                   mixins.RetrieveModelMixin,
                   viewsets.GenericViewSet):

Question 33

It appears that HyperlinkedModelSerializer do not agree with having a path namespace. In my application I made two changes.

# rootapp/urls.py
urlpatterns = [
    # path('api/', include('izzi.api.urls', namespace='api'))
    path('api/', include('izzi.api.urls')) # removed namespace
]

In the imported urls file

# app/urls.py
app_name = 'api' // removed the app_name

Hope this helps.

Question 34

I ran into the same error while I was following the DRF quickstart guide http://www.django-rest-framework.org/tutorial/quickstart/ and then attempting to browse to /users. I’ve done this setup many times before without problems.

My solution was not in the code but in replacing the database.

The difference between this install and the others before was when I created the local database.

This time I ran my

./manage.py migrate
./manage.py createsuperuser

immediately after running

virtualenv venv
. venv/bin/activate
pip install django
pip install djangorestframework

Instead of the exact order listed in the guide.

I suspected something wasn’t properly created in the DB. I didn’t care about my dev db so I deleted it and ran the ./manage.py migrate command once more, created a super user, browsed to /users and the error was gone.

Something was problematic with the order of operations in which I configured DRF and the db.

If you are using sqlite and are able to test changing to a fresh DB then it’s worth an attempt before you go dissecting all of your code.

Question 35

Bottle = serializers.PrimaryKeyRelatedField(read_only=True)

read_only allows you to represent the field without having to link it to another view of the model.

Question 36

I got that error on DRF 3.7.7 when a slug value was empty (equals to ”) in the database.

Question 37

I ran into this same issue and resolved it by adding generics.RetrieveAPIView as a base class to my viewset.

Question 38

I was stuck in this error for almost 2 hours:

ImproperlyConfigured at /api_users/users/1/ Could not resolve URL for hyperlinked relationship using view name “users-detail”. You may have failed to include the related model in your API, or incorrectly configured the lookup_field attribute on this field.

When I finally get the solution but I don’t understand why, so my code is:

#models.py
class Users(models.Model):
    id          = models.AutoField(primary_key=True)
    name        = models.CharField(max_length=50, blank=False, null=False)
    email       = models.EmailField(null=False, blank=False) 
    class Meta:
        verbose_name = "Usuario"
        verbose_name_plural = "Usuarios"

    def __str__(self):
        return str(self.name)


#serializers.py
class UserSerializer(serializers.HyperlinkedModelSerializer):
    class Meta:
        model = Users
        fields = (
            'id',
            'url',
            'name',        
            'email',       
            'description', 
            'active',      
            'age',         
            'some_date',   
            'timestamp',
            )
#views.py
class UserViewSet(viewsets.ModelViewSet):
    queryset = Users.objects.all()
    serializer_class = UserSerializer

#urls_api.py
router = routers.DefaultRouter()
router.register(r'users',UserViewSet, base_name='users')

urlpatterns = [ 
        url(r'^', include(router.urls)),
]

but in my main URLs, it was:

urlpatterns = [
    url(r'^admin/', admin.site.urls),
    #api users
    url(r'^api_users/', include('usersApi.users_urls', namespace='api')),

]

So to finally I resolve the problem erasing namespace:

urlpatterns = [
    url(r'^admin/', admin.site.urls),
    #api users
    url(r'^api_users/', include('usersApi.users_urls')),

]

And I finally resolve my problem, so any one can let me know why, bests.

Question 39

If you omit the fields ‘id’ and ‘url’ from your serializer you won’t have any problem. You can access to the posts by using the id that is returned in the json object anyways, which it makes it even easier to implement your frontend.

Question 40

I had the same problem , I think you should check your

get_absolute_url

object model’s method input value (**kwargs) title. and use exact field name in lookup_field

Question 41

I know that from Django 1.7 I don’t need to use South or any other migration system, so I am just using simple command python manage.py makemigrations

However, all I get is this error:

You are trying to add a non-nullable field 'new_field' to userprofile without a default;
we can't do that (the database needs something to populate existing rows).

Here is models.py:

class UserProfile(models.Model):
    user = models.OneToOneField(User)
    website = models.URLField(blank=True)
    new_field = models.CharField(max_length=140)

What are options?

Question 42

You need to provide a default value:

new_field = models.CharField(max_length=140, default='SOME STRING')

Question 43

If you are in early development cycle and don’t care about your current database data you can just remove it and then migrate. But first you need to clean migrations dir and remove its rows from table (django_migrations)

rm  your_app/migrations/*

rm db.sqlite3
python manage.py makemigrations
python manage.py migrate

Question 44

One option is to declare a default value for ‘new_field’:

new_field = models.CharField(max_length=140, default='DEFAULT VALUE')

another option is to declare ‘new_field’ as a nullable field:

new_field = models.CharField(max_length=140, null=True)

If you decide to accept ‘new_field’ as a nullable field you may want to accept ‘no input’ as valid input for ‘new_field’. Then you have to add the blank=True statement as well:

new_field = models.CharField(max_length=140, blank=True, null=True)

Even with null=True and/or blank=True you can add a default value if necessary:

new_field = models.CharField(max_length=140, default='DEFAULT VALUE', blank=True, null=True)

Question 45

If you are early into the development cycle you can try this –

Remove/comment that model and all its usages. Apply migrations. That would delete that model and then add the model again, run migrations and you have a clean model with the new field added.

Question 46

If “website” can be empty than new_field should also be set to be empty.

Now if you want to add logic on save where if new_field is empty to grab the value from “website” all you need to do is override the save function for your Model like this:

class UserProfile(models.Model):
    user = models.OneToOneField(User)
    website = models.URLField(blank=True, default='DEFAULT VALUE')
    new_field = models.CharField(max_length=140, blank=True, default='DEFAULT VALUE')

    def save(self, *args, **kwargs):
        if not self.new_field:
            # Setting the value of new_field with website's value
            self.new_field = self.website

        # Saving the object with the default save() function
        super(UserProfile, self).save(*args, **kwargs)

Question 47

In case anyone is setting a ForeignKey, you can just allow nullable fields without setting a default:

new_field = models.ForeignKey(model, null=True)

If you already have data stored within the database, you can also set a default value:

new_field = models.ForeignKey(model, default=<existing model id here>)

Question 48

You can’t add reference to table that have already data inside.
Change:

user = models.OneToOneField(User)

to:

user = models.OneToOneField(User, default = "")

do:

python manage.py makemigrations
python manage.py migrate

change again:

user = models.OneToOneField(User)

do migration again:

python manage.py makemigrations
python manage.py migrate

Question 49

In new_file add the boolean property null.

new_field = models.CharField(max_length=140, null=True)

after you run a ./manage.py syncdb for refresh the DB. and finally you run ./manage.py makemigrations and ./manage.py migrate

Question 50

You can use method from Django Doc from this page https://docs.djangoproject.com/en/1.8/ref/models/fields/#default

Create default and use it

def contact_default():
   return {"email": "to1@example.com"}

contact_info = JSONField("ContactInfo", default=contact_default)

Question 51

Do you already have database entries in the table UserProfile? If so, when you add new columns the DB doesn’t know what to set it to because it can’t be NULL. Therefore it asks you what you want to set those fields in the column new_fields to. I had to delete all the rows from this table to solve the problem.

(I know this was answered some time ago, but I just ran into this problem and this was my solution. Hopefully it will help anyone new that sees this)

Question 52

I honestly fount the best way to get around this was to just create another model with all the fields that you require and named slightly different. Run migrations. Delete unused model and run migrations again. Voila.

Question 53

If you are fine with truncating the table of the model in question, you can specify a one-off default value of None in the prompt. The migration will have superfluous default=None while your code has no default. It can be applied just fine because there’s no data in the table anymore which would require a default.

Question 54

I was early in my development cycle, so this may not work for everyone (but I don’t see why it wouldn’t).

I added blank=True, null=True to the columns where I was getting the error. Then I ran the python manage.py makemigrations command.

Immediately after running this command (and before running python manage.py migrate), I removed the blank=True, null=True from all the columns. Then I ran python manage.py makemigrations again. I was given an option to just change the columns myself, which I selected.

Then I ran python manage.py migrate and everything worked well!

Question 55

What Django actually says is:

Userprofile table has data in it and there might be new_field values which are null, but I do not know, so are you sure you want to mark property as non nullable, because if you do you might get an error if there are values with NULL

If you are sure that none of values in the userprofile table are NULL – fell free and ignore the warning.

The best practice in such cases would be to create a RunPython migration to handle empty values as it states in option 2

2) Ignore for now, and let me handle existing rows with NULL myself (e.g. because you added a RunPython or RunSQL operation to handle NULL values in a previous data migration)

In RunPython migration you have to find all UserProfile instances with empty new_field value and put a correct value there (or a default value as Django asks you to set in the model). You will get something like this:

# please keep in mind that new_value can be an empty string. You decide whether it is a correct value.
for profile in UserProfile.objects.filter(new_value__isnull=True).iterator():
    profile.new_value = calculate_value(profile)
    profile.save() # better to use batch save

Have fun!

Question 56

In models.py

class UserProfile(models.Model): user = models.OneToOneField(User) website = models.URLField(blank=True) new_field = models.CharField(max_length=140, default="some_value")

You need to add some values as default.

Question 57

If the SSH it gives you 2 options, choose number 1, and put “None”. Just that…for the moment.

Question 58

What is the equivalent of this SQL statement in django?

SELECT * FROM table_name WHERE string LIKE pattern;

How do I implement this in django? I tried

result = table.objects.filter( pattern in string )

But that did not work. How do i implement this?

Question 59

Use __contains or __icontains (case-insensitive):

result = table.objects.filter(string__contains='pattern')

The SQL equivalent is

SELECT ... WHERE string LIKE '%pattern%';

Question 60

contains and icontains mentioned by falsetru make queries like SELECT ... WHERE headline LIKE '%pattern%

Along with them, you might need these ones with similar behavior: startswith, istartswith, endswith, iendswith

making

SELECT ... WHERE headline LIKE 'pattern%

or

SELECT ... WHERE headline LIKE '%pattern

Question 61

result = table.objects.filter(string__icontains='pattern')

Case insensitive search for string in a field.

Question 62

In order to preserve the order of the words as in the sql LIKE ‘%pattern%’ statement I use iregex, for example:

qs = table.objects.filter(string__iregex=pattern.replace(' ', '.*'))

string methods are immutable so your pattern variable will not change and with .* you’ll be looking for 0 or more occurrences of any character but break lines.

By using the following to iterate over the pattern words:

qs = table.objects
for word in pattern.split(' '):
    qs = qs.filter(string__icontains=word)

the order of the words in your pattern will not be preserved, for some people that could work but in the case of trying to mimic the sql like statement I’ll use the first option.

Question 63

This can be done with Django’s custom lookups. I have made the lookup into a Django-like-lookup application. After installing it the __like lookup with the % and _ wildcards will be enabled.

All the necessary code in the application is:

from django.db.models import Lookup
from django.db.models.fields import Field


@Field.register_lookup
class Like(Lookup):
    lookup_name = 'like'

    def as_sql(self, compiler, connection):
        lhs, lhs_params = self.process_lhs(compiler, connection)
        rhs, rhs_params = self.process_rhs(compiler, connection)
        params = lhs_params + rhs_params
        return '%s LIKE %s' % (lhs, rhs), params

Question 64

I’m new to Python and Django.

I’m configuring a Django project using a PostgreSQL database engine backend, But I’m getting errors on each database operation. For example when I run manage.py syncdb, I’m getting:

C:\xampp\htdocs\djangodir>python manage.py syncdb
Traceback (most recent call last):
  File "manage.py", line 11, in <module>
    execute_manager(settings)
  File "C:\Python27\lib\site-packages\django\core\management\__init__.py", line
438, in execute_manager
    utility.execute()
  File "C:\Python27\lib\site-packages\django\core\management\__init__.py", line
379, in execute
    self.fetch_command(subcommand).run_from_argv(self.argv)
  File "C:\Python27\lib\site-packages\django\core\management\__init__.py", line
261, in fetch_command
    klass = load_command_class(app_name, subcommand)
  File "C:\Python27\lib\site-packages\django\core\management\__init__.py", line
67, in load_command_class
    module = import_module('%s.management.commands.%s' % (app_name, name))
  File "C:\Python27\lib\site-packages\django\utils\importlib.py", line 35, in im
port_module
    __import__(name)
  File "C:\Python27\lib\site-packages\django\core\management\commands\syncdb.py"
, line 7, in <module>
    from django.core.management.sql import custom_sql_for_model, emit_post_sync_
signal
  File "C:\Python27\lib\site-packages\django\core\management\sql.py", line 6, in
 <module>
    from django.db import models
  File "C:\Python27\lib\site-packages\django\db\__init__.py", line 77, in <modul
e>
    connection = connections[DEFAULT_DB_ALIAS]
  File "C:\Python27\lib\site-packages\django\db\utils.py", line 92, in __getitem
__
    backend = load_backend(db['ENGINE'])
  File "C:\Python27\lib\site-packages\django\db\utils.py", line 33, in load_back
end
    return import_module('.base', backend_name)
  File "C:\Python27\lib\site-packages\django\utils\importlib.py", line 35, in im
port_module
    __import__(name)
  File "C:\Python27\lib\site-packages\django\db\backends\postgresql\base.py", li
ne 23, in <module>
    raise ImproperlyConfigured("Error loading psycopg module: %s" % e)
django.core.exceptions.ImproperlyConfigured: Error loading psycopg module: No mo
dule named psycopg

Can someone give me a clue on what is going on?

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问题：您试图将非空字段’new_field’添加到用户配置文件中，而没有默认设置

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问题：Django查询中的“ LIKE”等效SQL

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安装

组态

Installation

Configuration

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问题：在Django Admin中调整字段大小