How to send a multipart form-data with requests in python

Question

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

User · Answer

You need to use the name attribute of the upload file that is in the HTML of the site  Example   autocomplete  off  name  image  gt    You see  name  image  gt   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

User · Answer

To clarify examples given above   quot You need to use the files parameter to send a multipart form POST request even when you do not need to upload any files  quot  files    won t work  unfortunately  You will need to put some dummy values in  e g  files   quot foo quot    quot bar quot    I came up against this when trying to upload files to Bitbucket s REST API and had to write this abomination to avoid the dreaded  quot Unsupported Media Type quot  error  url    quot https   my-bitbucket com rest api latest projects FOO repos bar browse foobar txt quot  payload     branch    master                content    text that will appear in my file               message    uploading directly from python   files     quot foo quot    quot bar quot   response   requests put url  data payload  files files    O

User · Answer

I m trying to send a request to URL server with request module in python 3  This works for me    - - coding  utf-8  -  import json  requests  URL SERVER TO POST DATA    quot URL to send POST request quot  HEADERS     quot Content-Type quot     quot multipart form-data  quot    def getPointsCC Function      file data            var1    None   quot valueOfYourVariable 1 quot           var2    None   quot valueOfYourVariable 2 quot          try      resElastic   requests post URL GET BALANCE  files file data      res   resElastic json     except Exception as e      print e     print  json dumps res  indent 4  sort keys True    getPointsCC Function    Where   URL SERVER TO POST DATA   Server where we going to send data HEADERS   Headers sended file data   Params sended

User · Answer

import requests   assume sending two files url    quot put ur url here quot  f1   open  quot file 1 path quot    rb   f2   open  quot file 2 path quot    rb   response   requests post url files   quot file1 name quot   f1   quot file2 name quot  f2   print response

User · Answer

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

User · Answer

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    gt  gt  gt  url    http   httpbin org post   gt  gt  gt  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    gt  gt  gt  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    gt  gt  gt  files     obvius session id    None   72c2b6f406cdabd578c5fd7598557c52     Content-Disposition  form-data  name  obvius session id  Content-Type  application octet-stream  72c2b6f406cdabd578c5fd7598557c52

User · Answer

Here is the simple code snippet to upload a single file with additional parameters using requests   url    https    lt file upload url gt   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

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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    gt  gt  gt  import requests  gt  gt  gt  response   requests post  http   httpbin org post   files dict foo  bar     gt  gt  gt  response status code 200   and httpbin org lets you know what headers you posted with  in response json   we have    gt  gt  gt  from pprint import pprint  gt  gt  gt  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    gt  gt  gt  files     foo    bar    gt  gt  gt  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--  gt  gt  gt  files     foo    None   bar     gt  gt  gt  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

User · Answer

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  lt Request gt                  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      9757  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

User · Answer

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

[python] How to send a "multipart/form-data" with requests in python?

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