标签归档:multipartform-data

如何在python中发送带有请求的“ multipart / form-data”?

问题:如何在python中发送带有请求的“ multipart / form-data”?

如何multipart/form-data在python中发送带有请求的请求?我了解如何发送文件,但是如何使用这种方法发送表单数据尚不清楚。

How to send a multipart/form-data with requests in python? How to send a file, I understand, but how to send the form data by this method can not understand.


回答 0

基本上,如果您指定files参数(字典),requests则将发送multipart/form-dataPOST而不是application/x-www-form-urlencodedPOST。您不限于在该词典中使用实际文件,但是:

>>> import requests
>>> response = requests.post('http://httpbin.org/post', files=dict(foo='bar'))
>>> response.status_code
200

httpbin.org可以让您知道您发布了哪些标题;在response.json()我们有:

>>> from pprint import pprint
>>> pprint(response.json()['headers'])
{'Accept': '*/*',
 'Accept-Encoding': 'gzip, deflate',
 'Connection': 'close',
 'Content-Length': '141',
 'Content-Type': 'multipart/form-data; '
                 'boundary=c7cbfdd911b4e720f1dd8f479c50bc7f',
 'Host': 'httpbin.org',
 'User-Agent': 'python-requests/2.21.0'}

更好的是,您可以通过使用元组而不是单个字符串或字节对象来进一步控制每个部分的文件名,内容类型和其他标题。元组应包含2到4个元素;文件名,内容,可选的内容类型以及其他标头的可选字典。

我将使用元组形式None作为文件名,以便filename="..."从那些部分的请求中删除参数:

>>> files = {'foo': 'bar'}
>>> print(requests.Request('POST', 'http://httpbin.org/post', files=files).prepare().body.decode('utf8'))
--bb3f05a247b43eede27a124ef8b968c5
Content-Disposition: form-data; name="foo"; filename="foo"

bar
--bb3f05a247b43eede27a124ef8b968c5--
>>> files = {'foo': (None, 'bar')}
>>> print(requests.Request('POST', 'http://httpbin.org/post', files=files).prepare().body.decode('utf8'))
--d5ca8c90a869c5ae31f70fa3ddb23c76
Content-Disposition: form-data; name="foo"

bar
--d5ca8c90a869c5ae31f70fa3ddb23c76--

files 如果需要排序和/或具有相同名称的多个字段,也可以是二值元组的列表:

requests.post(
    'http://requestb.in/xucj9exu',
    files=(
        ('foo', (None, 'bar')),
        ('foo', (None, 'baz')),
        ('spam', (None, 'eggs')),
    )
)

如果同时指定filesdata,然后它取决于价值data东西将被用于创建POST体。如果data是字符串,则仅使用它将;否则datafiles都使用和,data首先列出元素。

还有一个出色的requests-toolbelt项目,其中包括高级的Multipart支持。它采用与files参数格式相同的字段定义,但是与不同requests,它默认不设置文件名参数。另外,它可以从打开的文件对象流式传输请求,requests首先将在内存中构造请求主体:

from requests_toolbelt.multipart.encoder import MultipartEncoder

mp_encoder = MultipartEncoder(
    fields={
        'foo': 'bar',
        # plain file object, no filename or mime type produces a
        # Content-Disposition header with just the part name
        'spam': ('spam.txt', open('spam.txt', 'rb'), 'text/plain'),
    }
)
r = requests.post(
    'http://httpbin.org/post',
    data=mp_encoder,  # The MultipartEncoder is posted as data, don't use files=...!
    # The MultipartEncoder provides the content-type header with the boundary:
    headers={'Content-Type': mp_encoder.content_type}
)

字段遵循相同的约定;使用包含2到4个元素的元组来添加文件名,部分mime类型或额外的标头。不像files参数,没有试图找到一个默认filename值,如果你不使用的元组。

Basically, if you specify a files parameter (a dictionary), then requests will send a multipart/form-data POST instead of a application/x-www-form-urlencoded POST. You are not limited to using actual files in that dictionary, however:

>>> import requests
>>> response = requests.post('http://httpbin.org/post', files=dict(foo='bar'))
>>> response.status_code
200

and httpbin.org lets you know what headers you posted with; in response.json() we have:

>>> from pprint import pprint
>>> pprint(response.json()['headers'])
{'Accept': '*/*',
 'Accept-Encoding': 'gzip, deflate',
 'Connection': 'close',
 'Content-Length': '141',
 'Content-Type': 'multipart/form-data; '
                 'boundary=c7cbfdd911b4e720f1dd8f479c50bc7f',
 'Host': 'httpbin.org',
 'User-Agent': 'python-requests/2.21.0'}

Better still, you can further control the filename, content type and additional headers for each part by using a tuple instead of a single string or bytes object. The tuple is expected to contain between 2 and 4 elements; the filename, the content, optionally a content type, and an optional dictionary of further headers.

I’d use the tuple form with None as the filename, so that the filename="..." parameter is dropped from the request for those parts:

>>> files = {'foo': 'bar'}
>>> print(requests.Request('POST', 'http://httpbin.org/post', files=files).prepare().body.decode('utf8'))
--bb3f05a247b43eede27a124ef8b968c5
Content-Disposition: form-data; name="foo"; filename="foo"

bar
--bb3f05a247b43eede27a124ef8b968c5--
>>> files = {'foo': (None, 'bar')}
>>> print(requests.Request('POST', 'http://httpbin.org/post', files=files).prepare().body.decode('utf8'))
--d5ca8c90a869c5ae31f70fa3ddb23c76
Content-Disposition: form-data; name="foo"

bar
--d5ca8c90a869c5ae31f70fa3ddb23c76--

files can also be a list of two-value tuples, if you need ordering and/or multiple fields with the same name:

requests.post(
    'http://requestb.in/xucj9exu',
    files=(
        ('foo', (None, 'bar')),
        ('foo', (None, 'baz')),
        ('spam', (None, 'eggs')),
    )
)

If you specify both files and data, then it depends on the value of data what will be used to create the POST body. If data is a string, only it willl be used; otherwise both data and files are used, with the elements in data listed first.

There is also the excellent requests-toolbelt project, which includes advanced Multipart support. It takes field definitions in the same format as the files parameter, but unlike requests, it defaults to not setting a filename parameter. In addition, it can stream the request from open file objects, where requests will first construct the request body in memory:

from requests_toolbelt.multipart.encoder import MultipartEncoder

mp_encoder = MultipartEncoder(
    fields={
        'foo': 'bar',
        # plain file object, no filename or mime type produces a
        # Content-Disposition header with just the part name
        'spam': ('spam.txt', open('spam.txt', 'rb'), 'text/plain'),
    }
)
r = requests.post(
    'http://httpbin.org/post',
    data=mp_encoder,  # The MultipartEncoder is posted as data, don't use files=...!
    # The MultipartEncoder provides the content-type header with the boundary:
    headers={'Content-Type': mp_encoder.content_type}
)

Fields follow the same conventions; use a tuple with between 2 and 4 elements to add a filename, part mime-type or extra headers. Unlike the files parameter, no attempt is made to find a default filename value if you don’t use a tuple.


回答 1

自从编写了先前的答案以来,请求已更改。请查看Github上的bug线程以获取更多详细信息,并以示例的形式发表评论

简而言之,files参数使用a dict作为参数,键是表单字段的名称,值是字符串或2、3或4个长度的元组,如在请求中发布多部分编码文件一节中所述快速开始:

>>> url = 'http://httpbin.org/post'
>>> files = {'file': ('report.xls', open('report.xls', 'rb'), 'application/vnd.ms-excel', {'Expires': '0'})}

在上面,元组的组成如下:

(filename, data, content_type, headers)

如果该值只是一个字符串,则文件名将与键相同,如下所示:

>>> files = {'obvius_session_id': '72c2b6f406cdabd578c5fd7598557c52'}

Content-Disposition: form-data; name="obvius_session_id"; filename="obvius_session_id"
Content-Type: application/octet-stream

72c2b6f406cdabd578c5fd7598557c52

如果值是一个元组,并且第一个条目是Nonefilename属性,将不包括在内:

>>> files = {'obvius_session_id': (None, '72c2b6f406cdabd578c5fd7598557c52')}

Content-Disposition: form-data; name="obvius_session_id"
Content-Type: application/octet-stream

72c2b6f406cdabd578c5fd7598557c52

Since the previous answers were written, requests have changed. Have a look at the bug thread at Github for more detail and this comment for an example.

In short, the files parameter takes a dict with the key being the name of the form field and the value being either a string or a 2, 3 or 4-length tuple, as described in the section POST a Multipart-Encoded File in the requests quickstart:

>>> url = 'http://httpbin.org/post'
>>> files = {'file': ('report.xls', open('report.xls', 'rb'), 'application/vnd.ms-excel', {'Expires': '0'})}

In the above, the tuple is composed as follows:

(filename, data, content_type, headers)

If the value is just a string, the filename will be the same as the key, as in the following:

>>> files = {'obvius_session_id': '72c2b6f406cdabd578c5fd7598557c52'}

Content-Disposition: form-data; name="obvius_session_id"; filename="obvius_session_id"
Content-Type: application/octet-stream

72c2b6f406cdabd578c5fd7598557c52

If the value is a tuple and the first entry is None the filename property will not be included:

>>> files = {'obvius_session_id': (None, '72c2b6f406cdabd578c5fd7598557c52')}

Content-Disposition: form-data; name="obvius_session_id"
Content-Type: application/octet-stream

72c2b6f406cdabd578c5fd7598557c52

回答 2

即使不需要上传任何文件,也需要使用该files参数发送多部分表单POST请求。

从原始请求来源:

def request(method, url, **kwargs):
    """Constructs and sends a :class:`Request <Request>`.

    ...
    :param files: (optional) Dictionary of ``'name': file-like-objects``
        (or ``{'name': file-tuple}``) for multipart encoding upload.
        ``file-tuple`` can be a 2-tuple ``('filename', fileobj)``,
        3-tuple ``('filename', fileobj, 'content_type')``
        or a 4-tuple ``('filename', fileobj, 'content_type', custom_headers)``,
        where ``'content-type'`` is a string
        defining the content type of the given file
        and ``custom_headers`` a dict-like object 
        containing additional headers to add for the file.

相关部分是: file-tuple can be a2-tuple,。3-tupleor a4-tuple

基于上述内容,最简单的多部分表单请求包括要上传的文件和表单字段,如下所示:

multipart_form_data = {
    'file2': ('custom_file_name.zip', open('myfile.zip', 'rb')),
    'action': (None, 'store'),
    'path': (None, '/path1')
}

response = requests.post('https://httpbin.org/post', files=multipart_form_data)

print(response.content)

请注意,None作为纯文本字段的元组中的第一个参数-这是文件名字段的占位符,仅用于文件上传,但对于文本字段,传递None第一个参数是必需的,以便提交数据。

具有相同名称的多个字段

如果您需要发布多个具有相同名称的字段,那么您可以将有效负载定义为元组列表(或元组),而不是字典:

multipart_form_data = (
    ('file2', ('custom_file_name.zip', open('myfile.zip', 'rb'))),
    ('action', (None, 'store')),
    ('path', (None, '/path1')),
    ('path', (None, '/path2')),
    ('path', (None, '/path3')),
)

流请求API

如果上述API对您来说还不够Python,那么请考虑使用request工具带pip install requests_toolbelt),它是核心请求模块的扩展,该模块提供对文件上传流的支持以及MultipartEncoder(可以代替来使用)files,并且还可以您可以将有效负载定义为字典,元组或列表。

MultipartEncoder可以用于有或没有实际上传字段的多部分请求。必须将其分配给data参数。

import requests
from requests_toolbelt.multipart.encoder import MultipartEncoder

multipart_data = MultipartEncoder(
    fields={
            # a file upload field
            'file': ('file.zip', open('file.zip', 'rb'), 'text/plain')
            # plain text fields
            'field0': 'value0', 
            'field1': 'value1',
           }
    )

response = requests.post('http://httpbin.org/post', data=multipart_data,
                  headers={'Content-Type': multipart_data.content_type})

如果您需要发送多个具有相同名称的字段,或者表单字段的顺序很重要,则可以使用元组或列表代替字典:

multipart_data = MultipartEncoder(
    fields=(
            ('action', 'ingest'), 
            ('item', 'spam'),
            ('item', 'sausage'),
            ('item', 'eggs'),
           )
    )

You need to use the files parameter to send a multipart form POST request even when you do not need to upload any files.

From the original requests source:

def request(method, url, **kwargs):
    """Constructs and sends a :class:`Request <Request>`.

    ...
    :param files: (optional) Dictionary of ``'name': file-like-objects``
        (or ``{'name': file-tuple}``) for multipart encoding upload.
        ``file-tuple`` can be a 2-tuple ``('filename', fileobj)``,
        3-tuple ``('filename', fileobj, 'content_type')``
        or a 4-tuple ``('filename', fileobj, 'content_type', custom_headers)``,
        where ``'content-type'`` is a string
        defining the content type of the given file
        and ``custom_headers`` a dict-like object 
        containing additional headers to add for the file.

The relevant part is: file-tuple can be a2-tuple, 3-tupleor a4-tuple.

Based on the above, the simplest multipart form request that includes both files to upload and form fields will look like this:

multipart_form_data = {
    'file2': ('custom_file_name.zip', open('myfile.zip', 'rb')),
    'action': (None, 'store'),
    'path': (None, '/path1')
}

response = requests.post('https://httpbin.org/post', files=multipart_form_data)

print(response.content)

Note the None as the first argument in the tuple for plain text fields — this is a placeholder for the filename field which is only used for file uploads, but for text fields passing None as the first parameter is required in order for the data to be submitted.

Multiple fields with the same name

If you need to post multiple fields with the same name then instead of a dictionary you can define your payload as a list (or a tuple) of tuples:

multipart_form_data = (
    ('file2', ('custom_file_name.zip', open('myfile.zip', 'rb'))),
    ('action', (None, 'store')),
    ('path', (None, '/path1')),
    ('path', (None, '/path2')),
    ('path', (None, '/path3')),
)

Streaming requests API

If the above API is not pythonic enough for you, then consider using requests toolbelt (pip install requests_toolbelt) which is an extension of the core requests module that provides support for file upload streaming as well as the MultipartEncoder which can be used instead of files, and which also lets you define the payload as a dictionary, tuple or list.

MultipartEncoder can be used both for multipart requests with or without actual upload fields. It must be assigned to the data parameter.

import requests
from requests_toolbelt.multipart.encoder import MultipartEncoder

multipart_data = MultipartEncoder(
    fields={
            # a file upload field
            'file': ('file.zip', open('file.zip', 'rb'), 'text/plain')
            # plain text fields
            'field0': 'value0', 
            'field1': 'value1',
           }
    )

response = requests.post('http://httpbin.org/post', data=multipart_data,
                  headers={'Content-Type': multipart_data.content_type})

If you need to send multiple fields with the same name, or if the order of form fields is important, then a tuple or a list can be used instead of a dictionary:

multipart_data = MultipartEncoder(
    fields=(
            ('action', 'ingest'), 
            ('item', 'spam'),
            ('item', 'sausage'),
            ('item', 'eggs'),
           )
    )

回答 3

以下是使用请求上传带有其他参数的单个文件的简单代码段:

url = 'https://<file_upload_url>'
fp = '/Users/jainik/Desktop/data.csv'

files = {'file': open(fp, 'rb')}
payload = {'file_id': '1234'}

response = requests.put(url, files=files, data=payload, verify=False)

请注意,您不需要显式指定任何内容类型。

注意:想对以上答案之一发表评论,但由于声誉不佳而无法发表评论,因此在此处起草了一个新的答复。

Here is the simple code snippet to upload a single file with additional parameters using requests:

url = 'https://<file_upload_url>'
fp = '/Users/jainik/Desktop/data.csv'

files = {'file': open(fp, 'rb')}
payload = {'file_id': '1234'}

response = requests.put(url, files=files, data=payload, verify=False)

Please note that you don’t need to explicitly specify any content type.

NOTE: Wanted to comment on one of the above answers but could not because of low reputation so drafted a new response here.


回答 4

您需要使用name网站HTML中的上传文件的属性。例:

autocomplete="off" name="image">

看到了 name="image">吗 您可以在用于上传文件的网站的HTML中找到它。您需要使用它来上传文件Multipart/form-data

脚本:

import requests

site = 'https://prnt.sc/upload.php' # the site where you upload the file
filename = 'image.jpg'  # name example

在这里,在图片的位置,以HTML添加上传文件的名称

up = {'image':(filename, open(filename, 'rb'), "multipart/form-data")}

如果上传需要单击上传按钮,则可以这样使用:

data = {
     "Button" : "Submit",
}

然后开始请求

request = requests.post(site, files=up, data=data)

完成,文件成功上传

You need to use the name attribute of the upload file that is in the HTML of the site. Example:

autocomplete="off" name="image">

You see name="image">? You can find it in the HTML of a site for uploading the file. You need to use it to upload the file with Multipart/form-data

script:

import requests

site = 'https://prnt.sc/upload.php' # the site where you upload the file
filename = 'image.jpg'  # name example

Here, in the place of image, add the name of the upload file in HTML

up = {'image':(filename, open(filename, 'rb'), "multipart/form-data")}

If the upload requires to click the button for upload, you can use like that:

data = {
     "Button" : "Submit",
}

Then start the request

request = requests.post(site, files=up, data=data)

And done, file uploaded succesfully


回答 5

发送多部分/表单数据键和值

curl命令:

curl -X PUT http://127.0.0.1:8080/api/xxx ...
-H 'content-type: multipart/form-data; boundary=----xxx' \
-F taskStatus=1

python 请求-更复杂的POST请求

    updateTaskUrl = "http://127.0.0.1:8080/api/xxx"
    updateInfoDict = {
        "taskStatus": 1,
    }
    resp = requests.put(updateTaskUrl, data=updateInfoDict)

发送多部分/表单数据文件

curl命令:

curl -X POST http://127.0.0.1:8080/api/xxx ...
-H 'content-type: multipart/form-data; boundary=----xxx' \
-F file=@/Users/xxx.txt

python 请求-发布多部分编码的文件

    filePath = "/Users/xxx.txt"
    fileFp = open(filePath, 'rb')
    fileInfoDict = {
        "file": fileFp,
    }
    resp = requests.post(uploadResultUrl, files=fileInfoDict)

就这样。

Send multipart/form-data key and value

curl command:

curl -X PUT http://127.0.0.1:8080/api/xxx ...
-H 'content-type: multipart/form-data; boundary=----xxx' \
-F taskStatus=1

python requests – More complicated POST requests:

    updateTaskUrl = "http://127.0.0.1:8080/api/xxx"
    updateInfoDict = {
        "taskStatus": 1,
    }
    resp = requests.put(updateTaskUrl, data=updateInfoDict)

Send multipart/form-data file

curl command:

curl -X POST http://127.0.0.1:8080/api/xxx ...
-H 'content-type: multipart/form-data; boundary=----xxx' \
-F file=@/Users/xxx.txt

python requests – POST a Multipart-Encoded File:

    filePath = "/Users/xxx.txt"
    fileFp = open(filePath, 'rb')
    fileInfoDict = {
        "file": fileFp,
    }
    resp = requests.post(uploadResultUrl, files=fileInfoDict)

that’s all.


回答 6

这是您需要将一个大的单个文件作为多部分表单数据上传的python代码段。使用NodeJs Multer中间件在服务器端运行。

import requests
latest_file = 'path/to/file'
url = "http://httpbin.org/apiToUpload"
files = {'fieldName': open(latest_file, 'rb')}
r = requests.put(url, files=files)

对于服务器端,请在以下位置查看multer文档:https : //github.com/expressjs/multer, 此处字段single(’fieldName’)用于接受一个文件,如下所示:

var upload = multer().single('fieldName');

Here is the python snippet you need to upload one large single file as multipart formdata. With NodeJs Multer middleware running on the server side.

import requests
latest_file = 'path/to/file'
url = "http://httpbin.org/apiToUpload"
files = {'fieldName': open(latest_file, 'rb')}
r = requests.put(url, files=files)

For the server side please check the multer documentation at: https://github.com/expressjs/multer here the field single(‘fieldName’) is used to accept one single file, as in:

var upload = multer().single('fieldName');