标签归档:post

如何在Python中处理POST和GET变量?

问题:如何在Python中处理POST和GET变量?

在PHP中,您只能将其$_POST用于POST和$_GETGET(查询字符串)变量。Python中的等效功能是什么?

In PHP you can just use $_POST for POST and $_GET for GET (Query string) variables. What’s the equivalent in Python?


回答 0

假设您正在发布带有以下内容的html表单:

<input type="text" name="username">

如果使用原始cgi

import cgi
form = cgi.FieldStorage()
print form["username"]

如果使用DjangoPylonsFlaskPyramid

print request.GET['username'] # for GET form method
print request.POST['username'] # for POST form method

使用TurbogearsCherrypy

from cherrypy import request
print request.params['username']

Web.py

form = web.input()
print form.username

Werkzeug

print request.form['username']

如果使用Cherrypy或Turbogears,还可以直接使用参数定义处理程序函数:

def index(self, username):
    print username

Google App Engine

class SomeHandler(webapp2.RequestHandler):
    def post(self):
        name = self.request.get('username') # this will get the value from the field named username
        self.response.write(name) # this will write on the document

因此,您实际上必须选择这些框架之一。

suppose you’re posting a html form with this:

<input type="text" name="username">

If using raw cgi:

import cgi
form = cgi.FieldStorage()
print form["username"]

If using Django, Pylons, Flask or Pyramid:

print request.GET['username'] # for GET form method
print request.POST['username'] # for POST form method

Using Turbogears, Cherrypy:

from cherrypy import request
print request.params['username']

Web.py:

form = web.input()
print form.username

Werkzeug:

print request.form['username']

If using Cherrypy or Turbogears, you can also define your handler function taking a parameter directly:

def index(self, username):
    print username

Google App Engine:

class SomeHandler(webapp2.RequestHandler):
    def post(self):
        name = self.request.get('username') # this will get the value from the field named username
        self.response.write(name) # this will write on the document

So you really will have to choose one of those frameworks.


回答 1

我知道这是一个老问题。然而令人惊讶的是,没有给出好的答案。

首先,这个问题是完全有效的,而无需提及框架。CONTEXT是PHP语言的等效项。尽管有很多方法可以在Python中获取查询字符串参数,但是可以方便地填充框架变量。在PHP中,$_GET并且$_POST也方便变量。它们分别从QUERY_URI和php:// input解析。

在Python中,这些函数将是os.getenv('QUERY_STRING')sys.stdin.read()。记住要导入os和sys模块。

我们在这里必须小心使用“ CGI”一词,尤其是在谈论两种语言及其与Web服务器接口时的通用性时。1. CGI作为协议,定义了HTTP协议中的数据传输机制。2.可以将Python配置为在Apache中作为CGI脚本运行。3. Python中的CGI模块提供了一些便利功能。

由于HTTP协议与语言无关,并且Apache的CGI扩展也与语言无关,因此获取GET和POST参数仅应具有跨语言的语法差异。

这是填充GET字典的Python例程:

GET={}
args=os.getenv("QUERY_STRING").split('&')

for arg in args: 
    t=arg.split('=')
    if len(t)>1: k,v=arg.split('='); GET[k]=v

对于POST:

POST={}
args=sys.stdin.read().split('&')

for arg in args: 
    t=arg.split('=')
    if len(t)>1: k, v=arg.split('='); POST[k]=v

您现在可以按以下方式访问字段:

print GET.get('user_id')
print POST.get('user_name')

我还必须指出,CGI模块不能很好地工作。考虑以下HTTP请求:

POST / test.py?user_id=6

user_name=Bob&age=30

使用CGI.FieldStorage().getvalue('user_id')将导致空指针异常,因为该模块盲目检查POST数据,而忽略了POST请求也可以携带GET参数的事实。

I know this is an old question. Yet it’s surprising that no good answer was given.

First of all the question is completely valid without mentioning the framework. The CONTEXT is a PHP language equivalence. Although there are many ways to get the query string parameters in Python, the framework variables are just conveniently populated. In PHP, $_GET and $_POST are also convenience variables. They are parsed from QUERY_URI and php://input respectively.

In Python, these functions would be os.getenv('QUERY_STRING') and sys.stdin.read(). Remember to import os and sys modules.

We have to be careful with the word “CGI” here, especially when talking about two languages and their commonalities when interfacing with a web server. 1. CGI, as a protocol, defines the data transport mechanism in the HTTP protocol. 2. Python can be configured to run as a CGI-script in Apache. 3. The CGI module in Python offers some convenience functions.

Since the HTTP protocol is language-independent, and that Apache’s CGI extension is also language-independent, getting the GET and POST parameters should bear only syntax differences across languages.

Here’s the Python routine to populate a GET dictionary:

GET={}
args=os.getenv("QUERY_STRING").split('&')

for arg in args: 
    t=arg.split('=')
    if len(t)>1: k,v=arg.split('='); GET[k]=v

and for POST:

POST={}
args=sys.stdin.read().split('&')

for arg in args: 
    t=arg.split('=')
    if len(t)>1: k, v=arg.split('='); POST[k]=v

You can now access the fields as following:

print GET.get('user_id')
print POST.get('user_name')

I must also point out that the CGI module doesn’t work well. Consider this HTTP request:

POST / test.py?user_id=6

user_name=Bob&age=30

Using CGI.FieldStorage().getvalue('user_id') will cause a null pointer exception because the module blindly checks the POST data, ignoring the fact that a POST request can carry GET parameters too.


回答 2

我发现nosklo的答案非常广泛且有用!对于像我这样的人,他们可能会发现直接访问原始请求数据也很有用,我想添加一种方法:

import os, sys

# the query string, which contains the raw GET data
# (For example, for http://example.com/myscript.py?a=b&c=d&e
# this is "a=b&c=d&e")
os.getenv("QUERY_STRING")

# the raw POST data
sys.stdin.read()

I’ve found nosklo’s answer very extensive and useful! For those, like myself, who might find accessing the raw request data directly also useful, I would like to add the way to do that:

import os, sys

# the query string, which contains the raw GET data
# (For example, for http://example.com/myscript.py?a=b&c=d&e
# this is "a=b&c=d&e")
os.getenv("QUERY_STRING")

# the raw POST data
sys.stdin.read()

回答 3

它们存储在CGI fieldtorage对象中。

import cgi
form = cgi.FieldStorage()

print "The user entered %s" % form.getvalue("uservalue")

They are stored in the CGI fieldstorage object.

import cgi
form = cgi.FieldStorage()

print "The user entered %s" % form.getvalue("uservalue")

回答 4

它在某种程度上取决于您用作CGI框架的方式,但是在程序可访问的字典中可以找到它们。我会向您指出这些文档,但现在还没有到达python.org。但是mail.python.org上的此注释将为您提供第一个指针。查看CGI和URLLIB Python库以获取更多信息。

更新资料

好的,该链接无效。这是基本的wsgi参考

It somewhat depends on what you use as a CGI framework, but they are available in dictionaries accessible to the program. I’d point you to the docs, but I’m not getting through to python.org right now. But this note on mail.python.org will give you a first pointer. Look at the CGI and URLLIB Python libs for more.

Update

Okay, that link busted. Here’s the basic wsgi ref


回答 5

Python仅是一种语言,要获取GET和POST数据,您需要使用Python编写的Web框架或工具包。查理指出,Django是一个,cgi和urllib标准模块是另一个。也可以使用Turbogears,Pylons,CherryPy,web.py,mod_python,fastcgi等。

在Django中,您的视图函数会接收一个带有request.GET和request.POST的请求参数。其他框架将采取不同的方式。

Python is only a language, to get GET and POST data, you need a web framework or toolkit written in Python. Django is one, as Charlie points out, the cgi and urllib standard modules are others. Also available are Turbogears, Pylons, CherryPy, web.py, mod_python, fastcgi, etc, etc.

In Django, your view functions receive a request argument which has request.GET and request.POST. Other frameworks will do it differently.


如何以JSON格式发送POST请求?

问题:如何以JSON格式发送POST请求?

data = {
        'ids': [12, 3, 4, 5, 6 , ...]
    }
    urllib2.urlopen("http://abc.com/api/posts/create",urllib.urlencode(data))

我想发送POST请求,但是其中一个字段应该是数字列表。我怎样才能做到这一点 ?(JSON?)

data = {
        'ids': [12, 3, 4, 5, 6 , ...]
    }
    urllib2.urlopen("http://abc.com/api/posts/create",urllib.urlencode(data))

I want to send a POST request, but one of the fields should be a list of numbers. How can I do that ? (JSON?)


回答 0

如果您的服务器期望POST请求为json,那么您将需要添加标头,并为请求序列化数据…

Python 2.x

import json
import urllib2

data = {
        'ids': [12, 3, 4, 5, 6]
}

req = urllib2.Request('http://example.com/api/posts/create')
req.add_header('Content-Type', 'application/json')

response = urllib2.urlopen(req, json.dumps(data))

Python 3.x

https://stackoverflow.com/a/26876308/496445


如果不指定标题,它将是默认application/x-www-form-urlencoded类型。

If your server is expecting the POST request to be json, then you would need to add a header, and also serialize the data for your request…

Python 2.x

import json
import urllib2

data = {
        'ids': [12, 3, 4, 5, 6]
}

req = urllib2.Request('http://example.com/api/posts/create')
req.add_header('Content-Type', 'application/json')

response = urllib2.urlopen(req, json.dumps(data))

Python 3.x

https://stackoverflow.com/a/26876308/496445


If you don’t specify the header, it will be the default application/x-www-form-urlencoded type.


回答 1

我建议使用令人难以置信的requests模块。

http://docs.python-requests.org/zh-CN/v0.10.7/user/quickstart/#custom-headers

url = 'https://api.github.com/some/endpoint'
payload = {'some': 'data'}
headers = {'content-type': 'application/json'}

response = requests.post(url, data=json.dumps(payload), headers=headers)

I recommend using the incredible requests module.

http://docs.python-requests.org/en/v0.10.7/user/quickstart/#custom-headers

url = 'https://api.github.com/some/endpoint'
payload = {'some': 'data'}
headers = {'content-type': 'application/json'}

response = requests.post(url, data=json.dumps(payload), headers=headers)

回答 2

对于python 3.4.2,我发现以下将起作用:

import urllib.request
import json      

body = {'ids': [12, 14, 50]}  

myurl = "http://www.testmycode.com"
req = urllib.request.Request(myurl)
req.add_header('Content-Type', 'application/json; charset=utf-8')
jsondata = json.dumps(body)
jsondataasbytes = jsondata.encode('utf-8')   # needs to be bytes
req.add_header('Content-Length', len(jsondataasbytes))
print (jsondataasbytes)
response = urllib.request.urlopen(req, jsondataasbytes)

for python 3.4.2 I found the following will work:

import urllib.request
import json      

body = {'ids': [12, 14, 50]}  
myurl = "http://www.testmycode.com"

req = urllib.request.Request(myurl)
req.add_header('Content-Type', 'application/json; charset=utf-8')
jsondata = json.dumps(body)
jsondataasbytes = jsondata.encode('utf-8')   # needs to be bytes
req.add_header('Content-Length', len(jsondataasbytes))
response = urllib.request.urlopen(req, jsondataasbytes)

回答 3

这非常适合 Python 3.5,如果URL中包含查询字符串/参数值,

请求网址= https://bah2.com/ws/rest/v1/concept/
参数值= 21f6bb43-98a1-419d-8f0c-8133669e40ca

import requests

url = 'https://bahbah2.com/ws/rest/v1/concept/21f6bb43-98a1-419d-8f0c-8133669e40ca'
data = {"name": "Value"}
r = requests.post(url, auth=('username', 'password'), verify=False, json=data)
print(r.status_code)

This works perfect for Python 3.5, if the URL contains Query String / Parameter value,

Request URL = https://bah2.com/ws/rest/v1/concept/
Parameter value = 21f6bb43-98a1-419d-8f0c-8133669e40ca

import requests

url = 'https://bahbah2.com/ws/rest/v1/concept/21f6bb43-98a1-419d-8f0c-8133669e40ca'
data = {"name": "Value"}
r = requests.post(url, auth=('username', 'password'), verify=False, json=data)
print(r.status_code)

回答 4

您必须添加标题,否则会出现http 400错误。该代码在python2.6,centos5.4上运行良好

码:

    import urllib2,json

    url = 'http://www.google.com/someservice'
    postdata = {'key':'value'}

    req = urllib2.Request(url)
    req.add_header('Content-Type','application/json')
    data = json.dumps(postdata)

    response = urllib2.urlopen(req,data)

You have to add header,or you will get http 400 error. The code works well on python2.6,centos5.4

code:

    import urllib2,json

    url = 'http://www.google.com/someservice'
    postdata = {'key':'value'}

    req = urllib2.Request(url)
    req.add_header('Content-Type','application/json')
    data = json.dumps(postdata)

    response = urllib2.urlopen(req,data)

回答 5

这是一个如何使用Python标准库中的urllib.request对象的示例。

import urllib.request
import json
from pprint import pprint

url = "https://app.close.com/hackwithus/3d63efa04a08a9e0/"

values = {
    "first_name": "Vlad",
    "last_name": "Bezden",
    "urls": [
        "https://twitter.com/VladBezden",
        "https://github.com/vlad-bezden",
    ],
}


headers = {
    "Content-Type": "application/json",
    "Accept": "application/json",
}

data = json.dumps(values).encode("utf-8")
pprint(data)

try:
    req = urllib.request.Request(url, data, headers)
    with urllib.request.urlopen(req) as f:
        res = f.read()
    pprint(res.decode())
except Exception as e:
    pprint(e)

Here is an example of how to use urllib.request object from Python standard library.

import urllib.request
import json
from pprint import pprint

url = "https://app.close.com/hackwithus/3d63efa04a08a9e0/"

values = {
    "first_name": "Vlad",
    "last_name": "Bezden",
    "urls": [
        "https://twitter.com/VladBezden",
        "https://github.com/vlad-bezden",
    ],
}


headers = {
    "Content-Type": "application/json",
    "Accept": "application/json",
}

data = json.dumps(values).encode("utf-8")
pprint(data)

try:
    req = urllib.request.Request(url, data, headers)
    with urllib.request.urlopen(req) as f:
        res = f.read()
    pprint(res.decode())
except Exception as e:
    pprint(e)

回答 6

在最新的请求包中,您可以使用jsonin requests.post()方法来发送json dict,并且Content-Typein标头将设置为application/json。无需显式指定标头。

import requests

payload = {'key': 'value'}

requests.post(url, json=payload)

In the lastest requests package, you can use json parameter in requests.post() method to send a json dict, and the Content-Type in header will be set to application/json. There is no need to specify header explicitly.

import requests

payload = {'key': 'value'}

requests.post(url, json=payload)

回答 7

这对我使用api来说效果很好

import requests

data={'Id':id ,'name': name}
r = requests.post( url = 'https://apiurllink', data = data)

This one works fine for me with apis

import requests

data={'Id':id ,'name': name}
r = requests.post( url = 'https://apiurllink', data = data)

使用POST从Python脚本发送文件

问题:使用POST从Python脚本发送文件

有没有一种方法可以使用Python脚本中的POST发送文件?

Is there a way to send a file using POST from a Python script?


回答 0

来自:https : //requests.readthedocs.io/en/latest/user/quickstart/#post-a-multipart-encoded-file

通过请求,上传多部分编码的文件变得非常简单:

with open('report.xls', 'rb') as f:
    r = requests.post('http://httpbin.org/post', files={'report.xls': f})

而已。我不是在开玩笑-这是一行代码。文件已发送。让我们检查:

>>> r.text
{
  "origin": "179.13.100.4",
  "files": {
    "report.xls": "<censored...binary...data>"
  },
  "form": {},
  "url": "http://httpbin.org/post",
  "args": {},
  "headers": {
    "Content-Length": "3196",
    "Accept-Encoding": "identity, deflate, compress, gzip",
    "Accept": "*/*",
    "User-Agent": "python-requests/0.8.0",
    "Host": "httpbin.org:80",
    "Content-Type": "multipart/form-data; boundary=127.0.0.1.502.21746.1321131593.786.1"
  },
  "data": ""
}

From: https://requests.readthedocs.io/en/latest/user/quickstart/#post-a-multipart-encoded-file

Requests makes it very simple to upload Multipart-encoded files:

with open('report.xls', 'rb') as f:
    r = requests.post('http://httpbin.org/post', files={'report.xls': f})

That’s it. I’m not joking – this is one line of code. The file was sent. Let’s check:

>>> r.text
{
  "origin": "179.13.100.4",
  "files": {
    "report.xls": "<censored...binary...data>"
  },
  "form": {},
  "url": "http://httpbin.org/post",
  "args": {},
  "headers": {
    "Content-Length": "3196",
    "Accept-Encoding": "identity, deflate, compress, gzip",
    "Accept": "*/*",
    "User-Agent": "python-requests/0.8.0",
    "Host": "httpbin.org:80",
    "Content-Type": "multipart/form-data; boundary=127.0.0.1.502.21746.1321131593.786.1"
  },
  "data": ""
}

回答 1

是。您将使用urllib2模块,并使用multipart/form-data内容类型进行编码。这是一些入门示例代码-不仅仅是文件上传,但您应该能够阅读它并了解其工作方式:

user_agent = "image uploader"
default_message = "Image $current of $total"

import logging
import os
from os.path import abspath, isabs, isdir, isfile, join
import random
import string
import sys
import mimetypes
import urllib2
import httplib
import time
import re

def random_string (length):
    return ''.join (random.choice (string.letters) for ii in range (length + 1))

def encode_multipart_data (data, files):
    boundary = random_string (30)

    def get_content_type (filename):
        return mimetypes.guess_type (filename)[0] or 'application/octet-stream'

    def encode_field (field_name):
        return ('--' + boundary,
                'Content-Disposition: form-data; name="%s"' % field_name,
                '', str (data [field_name]))

    def encode_file (field_name):
        filename = files [field_name]
        return ('--' + boundary,
                'Content-Disposition: form-data; name="%s"; filename="%s"' % (field_name, filename),
                'Content-Type: %s' % get_content_type(filename),
                '', open (filename, 'rb').read ())

    lines = []
    for name in data:
        lines.extend (encode_field (name))
    for name in files:
        lines.extend (encode_file (name))
    lines.extend (('--%s--' % boundary, ''))
    body = '\r\n'.join (lines)

    headers = {'content-type': 'multipart/form-data; boundary=' + boundary,
               'content-length': str (len (body))}

    return body, headers

def send_post (url, data, files):
    req = urllib2.Request (url)
    connection = httplib.HTTPConnection (req.get_host ())
    connection.request ('POST', req.get_selector (),
                        *encode_multipart_data (data, files))
    response = connection.getresponse ()
    logging.debug ('response = %s', response.read ())
    logging.debug ('Code: %s %s', response.status, response.reason)

def make_upload_file (server, thread, delay = 15, message = None,
                      username = None, email = None, password = None):

    delay = max (int (delay or '0'), 15)

    def upload_file (path, current, total):
        assert isabs (path)
        assert isfile (path)

        logging.debug ('Uploading %r to %r', path, server)
        message_template = string.Template (message or default_message)

        data = {'MAX_FILE_SIZE': '3145728',
                'sub': '',
                'mode': 'regist',
                'com': message_template.safe_substitute (current = current, total = total),
                'resto': thread,
                'name': username or '',
                'email': email or '',
                'pwd': password or random_string (20),}
        files = {'upfile': path}

        send_post (server, data, files)

        logging.info ('Uploaded %r', path)
        rand_delay = random.randint (delay, delay + 5)
        logging.debug ('Sleeping for %.2f seconds------------------------------\n\n', rand_delay)
        time.sleep (rand_delay)

    return upload_file

def upload_directory (path, upload_file):
    assert isabs (path)
    assert isdir (path)

    matching_filenames = []
    file_matcher = re.compile (r'\.(?:jpe?g|gif|png)$', re.IGNORECASE)

    for dirpath, dirnames, filenames in os.walk (path):
        for name in filenames:
            file_path = join (dirpath, name)
            logging.debug ('Testing file_path %r', file_path)
            if file_matcher.search (file_path):
                matching_filenames.append (file_path)
            else:
                logging.info ('Ignoring non-image file %r', path)

    total_count = len (matching_filenames)
    for index, file_path in enumerate (matching_filenames):
        upload_file (file_path, index + 1, total_count)

def run_upload (options, paths):
    upload_file = make_upload_file (**options)

    for arg in paths:
        path = abspath (arg)
        if isdir (path):
            upload_directory (path, upload_file)
        elif isfile (path):
            upload_file (path)
        else:
            logging.error ('No such path: %r' % path)

    logging.info ('Done!')

Yes. You’d use the urllib2 module, and encode using the multipart/form-data content type. Here is some sample code to get you started — it’s a bit more than just file uploading, but you should be able to read through it and see how it works:

user_agent = "image uploader"
default_message = "Image $current of $total"

import logging
import os
from os.path import abspath, isabs, isdir, isfile, join
import random
import string
import sys
import mimetypes
import urllib2
import httplib
import time
import re

def random_string (length):
    return ''.join (random.choice (string.letters) for ii in range (length + 1))

def encode_multipart_data (data, files):
    boundary = random_string (30)

    def get_content_type (filename):
        return mimetypes.guess_type (filename)[0] or 'application/octet-stream'

    def encode_field (field_name):
        return ('--' + boundary,
                'Content-Disposition: form-data; name="%s"' % field_name,
                '', str (data [field_name]))

    def encode_file (field_name):
        filename = files [field_name]
        return ('--' + boundary,
                'Content-Disposition: form-data; name="%s"; filename="%s"' % (field_name, filename),
                'Content-Type: %s' % get_content_type(filename),
                '', open (filename, 'rb').read ())

    lines = []
    for name in data:
        lines.extend (encode_field (name))
    for name in files:
        lines.extend (encode_file (name))
    lines.extend (('--%s--' % boundary, ''))
    body = '\r\n'.join (lines)

    headers = {'content-type': 'multipart/form-data; boundary=' + boundary,
               'content-length': str (len (body))}

    return body, headers

def send_post (url, data, files):
    req = urllib2.Request (url)
    connection = httplib.HTTPConnection (req.get_host ())
    connection.request ('POST', req.get_selector (),
                        *encode_multipart_data (data, files))
    response = connection.getresponse ()
    logging.debug ('response = %s', response.read ())
    logging.debug ('Code: %s %s', response.status, response.reason)

def make_upload_file (server, thread, delay = 15, message = None,
                      username = None, email = None, password = None):

    delay = max (int (delay or '0'), 15)

    def upload_file (path, current, total):
        assert isabs (path)
        assert isfile (path)

        logging.debug ('Uploading %r to %r', path, server)
        message_template = string.Template (message or default_message)

        data = {'MAX_FILE_SIZE': '3145728',
                'sub': '',
                'mode': 'regist',
                'com': message_template.safe_substitute (current = current, total = total),
                'resto': thread,
                'name': username or '',
                'email': email or '',
                'pwd': password or random_string (20),}
        files = {'upfile': path}

        send_post (server, data, files)

        logging.info ('Uploaded %r', path)
        rand_delay = random.randint (delay, delay + 5)
        logging.debug ('Sleeping for %.2f seconds------------------------------\n\n', rand_delay)
        time.sleep (rand_delay)

    return upload_file

def upload_directory (path, upload_file):
    assert isabs (path)
    assert isdir (path)

    matching_filenames = []
    file_matcher = re.compile (r'\.(?:jpe?g|gif|png)$', re.IGNORECASE)

    for dirpath, dirnames, filenames in os.walk (path):
        for name in filenames:
            file_path = join (dirpath, name)
            logging.debug ('Testing file_path %r', file_path)
            if file_matcher.search (file_path):
                matching_filenames.append (file_path)
            else:
                logging.info ('Ignoring non-image file %r', path)

    total_count = len (matching_filenames)
    for index, file_path in enumerate (matching_filenames):
        upload_file (file_path, index + 1, total_count)

def run_upload (options, paths):
    upload_file = make_upload_file (**options)

    for arg in paths:
        path = abspath (arg)
        if isdir (path):
            upload_directory (path, upload_file)
        elif isfile (path):
            upload_file (path)
        else:
            logging.error ('No such path: %r' % path)

    logging.info ('Done!')

回答 2

阻止您直接在文件对象上直接使用urlopen的唯一原因是内置文件对象缺少len定义。一种简单的方法是创建一个子类,该子类为urlopen提供正确的文件。我还修改了下面文件中的Content-Type标头。

import os
import urllib2
class EnhancedFile(file):
    def __init__(self, *args, **keyws):
        file.__init__(self, *args, **keyws)

    def __len__(self):
        return int(os.fstat(self.fileno())[6])

theFile = EnhancedFile('a.xml', 'r')
theUrl = "http://example.com/abcde"
theHeaders= {'Content-Type': 'text/xml'}

theRequest = urllib2.Request(theUrl, theFile, theHeaders)

response = urllib2.urlopen(theRequest)

theFile.close()


for line in response:
    print line

The only thing that stops you from using urlopen directly on a file object is the fact that the builtin file object lacks a len definition. A simple way is to create a subclass, which provides urlopen with the correct file. I have also modified the Content-Type header in the file below.

import os
import urllib2
class EnhancedFile(file):
    def __init__(self, *args, **keyws):
        file.__init__(self, *args, **keyws)

    def __len__(self):
        return int(os.fstat(self.fileno())[6])

theFile = EnhancedFile('a.xml', 'r')
theUrl = "http://example.com/abcde"
theHeaders= {'Content-Type': 'text/xml'}

theRequest = urllib2.Request(theUrl, theFile, theHeaders)

response = urllib2.urlopen(theRequest)

theFile.close()


for line in response:
    print line

回答 3

看起来python请求无法处理非常大的多部分文件。

该文档建议您仔细阅读requests-toolbelt

这是他们文档中的相关页面

Looks like python requests does not handle extremely large multi-part files.

The documentation recommends you look into requests-toolbelt.

Here’s the pertinent page from their documentation.


回答 4

克里斯·阿特利(Chris Atlee)的海报库在此方面非常有效(尤其是便捷功能poster.encode.multipart_encode())。另外,它支持流大文件而无需将整个文件加载到内存中。另请参阅Python问题3244

Chris Atlee’s poster library works really well for this (particularly the convenience function poster.encode.multipart_encode()). As a bonus, it supports streaming of large files without loading an entire file into memory. See also Python issue 3244.


回答 5

我正在尝试测试Django Rest API及其对我的工作:

def test_upload_file(self):
        filename = "/Users/Ranvijay/tests/test_price_matrix.csv"
        data = {'file': open(filename, 'rb')}
        client = APIClient()
        # client.credentials(HTTP_AUTHORIZATION='Token ' + token.key)
        response = client.post(reverse('price-matrix-csv'), data, format='multipart')

        print response
        self.assertEqual(response.status_code, status.HTTP_200_OK)

I am trying to test django rest api and its working for me:

def test_upload_file(self):
        filename = "/Users/Ranvijay/tests/test_price_matrix.csv"
        data = {'file': open(filename, 'rb')}
        client = APIClient()
        # client.credentials(HTTP_AUTHORIZATION='Token ' + token.key)
        response = client.post(reverse('price-matrix-csv'), data, format='multipart')

        print response
        self.assertEqual(response.status_code, status.HTTP_200_OK)

回答 6

您可能还想看看带有示例的httplib2。我发现使用httplib2比使用内置HTTP模块更为简洁。

You may also want to have a look at httplib2, with examples. I find using httplib2 is more concise than using the built-in HTTP modules.


回答 7

def visit_v2(device_code, camera_code):
    image1 = MultipartParam.from_file("files", "/home/yuzx/1.txt")
    image2 = MultipartParam.from_file("files", "/home/yuzx/2.txt")
    datagen, headers = multipart_encode([('device_code', device_code), ('position', 3), ('person_data', person_data), image1, image2])
    print "".join(datagen)
    if server_port == 80:
        port_str = ""
    else:
        port_str = ":%s" % (server_port,)
    url_str = "http://" + server_ip + port_str + "/adopen/device/visit_v2"
    headers['nothing'] = 'nothing'
    request = urllib2.Request(url_str, datagen, headers)
    try:
        response = urllib2.urlopen(request)
        resp = response.read()
        print "http_status =", response.code
        result = json.loads(resp)
        print resp
        return result
    except urllib2.HTTPError, e:
        print "http_status =", e.code
        print e.read()
def visit_v2(device_code, camera_code):
    image1 = MultipartParam.from_file("files", "/home/yuzx/1.txt")
    image2 = MultipartParam.from_file("files", "/home/yuzx/2.txt")
    datagen, headers = multipart_encode([('device_code', device_code), ('position', 3), ('person_data', person_data), image1, image2])
    print "".join(datagen)
    if server_port == 80:
        port_str = ""
    else:
        port_str = ":%s" % (server_port,)
    url_str = "http://" + server_ip + port_str + "/adopen/device/visit_v2"
    headers['nothing'] = 'nothing'
    request = urllib2.Request(url_str, datagen, headers)
    try:
        response = urllib2.urlopen(request)
        resp = response.read()
        print "http_status =", response.code
        result = json.loads(resp)
        print resp
        return result
    except urllib2.HTTPError, e:
        print "http_status =", e.code
        print e.read()

如何在Flask中获取POST JSON?

问题:如何在Flask中获取POST JSON?

我正在尝试使用Flask构建一个简单的API,现在我想在其中读取一些POSTed JSON。我使用Postman Chrome扩展程序执行POST,而我发布的JSON就是{"text":"lalala"}。我尝试使用以下方法读取JSON:

@app.route('/api/add_message/<uuid>', methods=['GET', 'POST'])
def add_message(uuid):
    content = request.json
    print content
    return uuid

在浏览器上,它可以正确返回我放入GET中的UUID,但是在控制台上,它只是打印出来None(我希望它可以在其中打印出来{"text":"lalala"}。有人知道我如何从Flask方法中获取发布的JSON吗?

I’m trying to build a simple API using Flask, in which I now want to read some POSTed JSON. I do the POST with the Postman Chrome extension, and the JSON I POST is simply {"text":"lalala"}. I try to read the JSON using the following method:

@app.route('/api/add_message/<uuid>', methods=['GET', 'POST'])
def add_message(uuid):
    content = request.json
    print content
    return uuid

On the browser it correctly returns the UUID I put in the GET, but on the console, it just prints out None (where I expect it to print out the {"text":"lalala"}. Does anybody know how I can get the posted JSON from within the Flask method?


回答 0

首先,该.json属性是一个委托给request.get_json()method的属性,该属性记录了为什么None在此处看到。

您需要将请求内容类型设置为,application/json以使.json属性和.get_json()方法(不带参数)起作用,None否则两者都会产生这种情况。请参阅Flask Request文档

如果mimetype表示JSON(包含application / json,请参见is_json()),则它将包含已解析的JSON数据,否则为None

您可以request.get_json()通过向其传递force=True关键字参数来告知跳过内容类型要求。

请注意,如果此时引发异常(可能导致400 Bad Request响应),则您的JSON 数据无效。它在某种程度上畸形;您可能需要使用JSON验证程序进行检查。

First of all, the .json attribute is a property that delegates to the request.get_json() method, which documents why you see None here.

You need to set the request content type to application/json for the .json property and .get_json() method (with no arguments) to work as either will produce None otherwise. See the Flask Request documentation:

This will contain the parsed JSON data if the mimetype indicates JSON (application/json, see is_json()), otherwise it will be None.

You can tell request.get_json() to skip the content type requirement by passing it the force=True keyword argument.

Note that if an exception is raised at this point (possibly resulting in a 400 Bad Request response), your JSON data is invalid. It is in some way malformed; you may want to check it with a JSON validator.


回答 1

作为参考,以下是有关如何从Python客户端发送json的完整代码:

import requests
res = requests.post('http://localhost:5000/api/add_message/1234', json={"mytext":"lalala"})
if res.ok:
    print res.json()

“ json =“输入将自动设置内容类型,如下所述:使用Python请求发布JSON

以上客户端将使用此服务器端代码:

from flask import Flask, request, jsonify
app = Flask(__name__)

@app.route('/api/add_message/<uuid>', methods=['GET', 'POST'])
def add_message(uuid):
    content = request.json
    print content['mytext']
    return jsonify({"uuid":uuid})

if __name__ == '__main__':
    app.run(host= '0.0.0.0',debug=True)

For reference, here’s complete code for how to send json from a Python client:

import requests
res = requests.post('http://localhost:5000/api/add_message/1234', json={"mytext":"lalala"})
if res.ok:
    print res.json()

The “json=” input will automatically set the content-type, as discussed here: Post JSON using Python Requests

And the above client will work with this server-side code:

from flask import Flask, request, jsonify
app = Flask(__name__)

@app.route('/api/add_message/<uuid>', methods=['GET', 'POST'])
def add_message(uuid):
    content = request.json
    print content['mytext']
    return jsonify({"uuid":uuid})

if __name__ == '__main__':
    app.run(host= '0.0.0.0',debug=True)

回答 2

这就是我要做的方法,应该是

@app.route('/api/add_message/<uuid>', methods=['GET', 'POST'])
def add_message(uuid):
    content = request.get_json(silent=True)
    # print(content) # Do your processing
    return uuid

随着silent=True集,该get_json功能将尝试检索JSON的身体的时候默默的失败。默认情况下,此设置为False。如果您始终希望使用json正文(而不是可选),请将其保留为silent=False

设置force=True将忽略request.headers.get('Content-Type') == 'application/json'烧瓶对您的 检查。默认情况下,它也设置为False

请参见烧瓶文档

我强烈建议您离开force=False并让客户端发送Content-Type标头以使其更加明确。

希望这可以帮助!

This is the way I would do it and it should be

@app.route('/api/add_message/<uuid>', methods=['GET', 'POST'])
def add_message(uuid):
    content = request.get_json(silent=True)
    # print(content) # Do your processing
    return uuid

With silent=True set, the get_json function will fail silently when trying to retrieve the json body. By default this is set to False. If you are always expecting a json body (not optionally), leave it as silent=False.

Setting force=True will ignore the request.headers.get('Content-Type') == 'application/json' check that flask does for you. By default this is also set to False.

See flask documentation.

I would strongly recommend leaving force=False and make the client send the Content-Type header to make it more explicit.

Hope this helps!


回答 3

假设您已发布具有application/json内容类型的有效JSON ,request.json将具有已解析的JSON数据。

from flask import Flask, request, jsonify

app = Flask(__name__)


@app.route('/echo', methods=['POST'])
def hello():
   return jsonify(request.json)

Assuming you’ve posted valid JSON with the application/json content type, request.json will have the parsed JSON data.

from flask import Flask, request, jsonify

app = Flask(__name__)


@app.route('/echo', methods=['POST'])
def hello():
   return jsonify(request.json)

回答 4

对于所有问题都是来自ajax调用的人,这里有一个完整的示例:

Ajax调用:这里的关键是使用a dict然后JSON.stringify

    var dict = {username : "username" , password:"password"};

    $.ajax({
        type: "POST", 
        url: "http://127.0.0.1:5000/", //localhost Flask
        data : JSON.stringify(dict),
        contentType: "application/json",
    });

在服务器端:

from flask import Flask
from flask import request
import json

app = Flask(__name__)

@app.route("/",  methods = ['POST'])
def hello():
    print(request.get_json())
    return json.dumps({'success':True}), 200, {'ContentType':'application/json'} 

if __name__ == "__main__":
    app.run()

For all those whose issue was from the ajax call, here is a full example :

Ajax call : the key here is to use a dict and then JSON.stringify

    var dict = {username : "username" , password:"password"};

    $.ajax({
        type: "POST", 
        url: "http://127.0.0.1:5000/", //localhost Flask
        data : JSON.stringify(dict),
        contentType: "application/json",
    });

And on server side :

from flask import Flask
from flask import request
import json

app = Flask(__name__)

@app.route("/",  methods = ['POST'])
def hello():
    print(request.get_json())
    return json.dumps({'success':True}), 200, {'ContentType':'application/json'} 

if __name__ == "__main__":
    app.run()

回答 5

给出另一种方法。

from flask import Flask, jsonify, request
app = Flask(__name__)

@app.route('/service', methods=['POST'])
def service():
    data = json.loads(request.data)
    text = data.get("text",None)
    if text is None:
        return jsonify({"message":"text not found"})
    else:
        return jsonify(data)

if __name__ == '__main__':
    app.run(host= '0.0.0.0',debug=True)

To give another approach.

from flask import Flask, jsonify, request
app = Flask(__name__)

@app.route('/service', methods=['POST'])
def service():
    data = json.loads(request.data)
    text = data.get("text",None)
    if text is None:
        return jsonify({"message":"text not found"})
    else:
        return jsonify(data)

if __name__ == '__main__':
    app.run(host= '0.0.0.0',debug=True)

回答 6

假设您发布了有效的JSON,

@app.route('/api/add_message/<uuid>', methods=['GET', 'POST'])
def add_message(uuid):
    content = request.json
    print content['uuid']
    # Return data as JSON
    return jsonify(content)

Assuming that you have posted valid JSON,

@app.route('/api/add_message/<uuid>', methods=['GET', 'POST'])
def add_message(uuid):
    content = request.json
    print content['uuid']
    # Return data as JSON
    return jsonify(content)

回答 7

尝试使用力参数…

request.get_json(force = True)

Try to use force parameter…

request.get_json(force = True)