将views.py拆分为多个文件

My views.py has become too big and it’s hard to find the right view.

How do I split it in several files and then import it? Does it involve any speed loss?

Can I do the same with models.py?

In Django everything is a Python module (*.py). You can create a view folder with an __init__.py inside and you still will be able to import your views, because this also implements a Python module. But an example would be better.

Your original views.py might look like this :

def view1(arg):
    pass

def view2(arg):
   pass

With the following folder/file structure it will work the same :

views/
   __init__.py
   viewsa.py
   viewsb.py

viewsa.py :

def view1(arg):
    pass

viewsb.py :

def view2(arg):
    pass

__init__.py :

from viewsa import view1
from viewsb import view2

The quick explanation would be: when you write from views import view1 Python will look for view1 in

1. views.py, which is what happens in the first (original) case

2. views/__init__.py, which is what happens in the second case. Here, __init__.py is able to provide the view1 method because it imports it.

With this kind of solution, you might have no need to change import or urlpatterns arguments in urls.py

If you have many methods in each new view file, you might find it useful to make the imports in views/__init__.py use *, like this:

from viewsa import *
from viewsb import *

I actually don’t know about speed issues (but I doubt there are any).

For Models it might be a bit difficult.

I’ve had to do this before (for clarities sake)

The way I did this was to create a views directory, then, in that, create a file called __init__.py

Now, when you’re calling in your urls.py, you simply need to add another part

For example, previously, you may have called:-

url(r'^calendar/(?P<year>\d\d\d\d)/$', 'myproject.calendar.views.year')
url(r'^calendar/(?P<year>\d\d\d\d)/(?P<user>[a-z]+)/$', 'myproject.calendar.views.year_by_user')

You can now call something along the lines of

url(r'^calendar/(?P<year>\d\d\d\d)/$', 'myproject.calendar.views.year.index')
url(r'^calendar/(?P<year>\d\d\d\d)/(?P<user>[a-z]+)/$', 'myproject.calendar.views.year.user')

This is, of course, assuming that you had views/year.py containing the functions index and user ;)

Basically, you can put your code, whereever you wish. Just make sure, you change the import statements accordingly, e.g. for the views in the urls.py.

Not knowing your actual code its hard to suggest something meaningful. Maybe you can use some kind of filename prefix, e.g. views_helper.py, views_fancy.py, views_that_are_not_so_often_used.py or so …

Another option would be to create a views directory with an __init__.py, where you import all subviews. If you have a need for a large number of files, you can create more nested subviews as your views grow …

Just for sharing, I had a bit of issues with Vincent Demeester’s answer. Everything is fine except in init.py file, I have to write in this way:

__init__.py:

from .viewsa import *
from .viewsb import *

This way I still don’t need to change my import method in urls.py. I am on Python 3.6.1 and Django 1.11.4.

Simple answer: Yes.

Best is to make a directory called views and then in your urls.py do:

import views
...
url(r'^classroom$', views.school.klass, name="classroom"),

I split almost all views in my apps into a views folder (with an init.py of course). I do not, however, import all of the subviews in the init.py like some of the answers have suggested. It seems to work just fine.

Since Django just expects a view to be a callable object, you can put then wherever you like in your PYTHONPATH. So you could for instance just make a new package myapp.views and put views into multiple modules there. You will naturally have to update your urls.py and other modules that reference these view callables.

I’ve been playing with putting this in my init.py:

import os

currPath = os.path.realpath(os.path.dirname(__file__))

dirFiles = []
for root, dirs, files in os.walk(currPath):
    for name in files:
        if name.endswith('.py') and not name.startswith('_'): 
            dirFiles.append(name.strip('.py'))

for f in dirFiles:
    exec("from %s import %s" % (f,f))

I’m still new to python, so I’m still looking at what effect it has on speed/security/ease of use.

Suppose if you have a file named: password_generator.py then inside views.py add: from password_generator import *

Then you can call that module’s function from views.py.

Vincent Demeester‘s answer is superb! but for me addicted‘s answer worked like a charm. I faced difficulties in migrating database. The error indicates the line where the first model is imported and says could not recognize my app module. Searched a lot but could not find a solution but later on I imported the model like this:

from ..models import ModelName

It worked!!