Uploading multiple files in a single request using python requests module

Question

The Python requests module provides good documentation on how to upload a single file in a single request:

 files = {'file': open('report.xls', 'rb')}

I tried extending that example by using this code in an attempt to upload multiple files:

 files = {'file': [open('report.xls', 'rb'), open('report2.xls, 'rb')]}

but it resulted in this error:

 File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/urllib.py",      line 1052, in splittype
 match = _typeprog.match(url)
 TypeError: expected string or buffer

Is it possible to upload a list of files in a single request using this module, and how?

Answer 1

Multiple files with different key values can be uploaded by adding multiple dictionary entries:

files = {'file1': open('report.xls', 'rb'), 'file2': open('otherthing.txt', 'rb')}
r = requests.post('http://httpbin.org/post', files=files)

Answer 2

To upload a list of files with the same key value in a single request, you can create a list of tuples with the first item in each tuple as the key value and the file object as the second:

files = [('file', open('report.xls', 'rb')), ('file', open('report2.xls', 'rb'))]

Answer 3

The documentation contains a clear answer.

Quoted:

You can send multiple files in one request. For example, suppose you want to upload image files to an HTML form with a multiple file field ‘images’:

To do that, just set files to a list of tuples of (form_field_name, file_info):

url = 'http://httpbin.org/post'
multiple_files = [('images', ('foo.png', open('foo.png', 'rb'), 'image/png')),
                      ('images', ('bar.png', open('bar.png', 'rb'), 'image/png'))]
r = requests.post(url, files=multiple_files)
r.text

# {
#  ...
#  'files': {'images': 'data:image/png;base64,iVBORw ....'}
#  'Content-Type': 'multipart/form-data; boundary=3131623adb2043caaeb5538cc7aa0b3a',
#  ...
# }

Answer 4

You need to create a file list to upload multiple images:

file_list = [  
       ('Key_here', ('file_name1.jpg', open('file_path1.jpg', 'rb'), 'image/png')),
       ('key_here', ('file_name2.jpg', open('file_path2.jpg', 'rb'), 'image/png'))
   ]

r = requests.post(url, files=file_list)

If you want to send files on the same key you need to keep the key same for each element, and for a different key just change the keys.

Source : https://stackabuse.com/the-python-requests-module/

Answer 5

I'm a little bit confused, but directly opening file in request (however same is written in official requests guide) is not so "safe".

Just try:

import os
import requests
file_path = "/home/user_folder/somefile.txt"
files = {'somefile': open(file_path, 'rb')}
r = requests.post('http://httpbin.org/post', files=files)

Yes, all will be ok, but:

os.rename(file_path, file_path)

And you will get:

PermissionError:The process cannot access the file because it is being used by another process

Please, correct me if I'm not right, but it seems that file is still opened and I do not know any way to close it.

Instead of this I use:

import os
import requests
#let it be folder with files to upload
folder = "/home/user_folder/"
#dict for files
upload_list = []
for files in os.listdir(folder):
    with open("{folder}{name}".format(folder=folder, name=files), "rb") as data:
        upload_list.append(files, data.read())
r = request.post("https://httpbin.org/post", files=upload_list)
#trying to rename uploaded files now
for files in os.listdir(folder):
    os.rename("{folder}{name}".format(folder=folder, name=files), "{folder}{name}".format(folder=folder, name=files))

Now we do not get errors, so I recommend to use this way to upload multiple files or you could receive some errors. Hope this answer well help somebody and save priceless time.

Answer 6

If you have multiple files in a python list, you can use eval() in a comprehension to loop over the files in the requests post files parameter.

file_list = ['001.jpg', '002.jpg', '003.jpg']
files=[eval(f'("inline", open("{file}", "rb"))') for file in file_list ]

requests.post(
        url=url,
        files=files
)

Answer 7

In my case uploading all the images which are inside the folder just adding key with index

e.g. key = 'images' to e.g. 'images[0]' in the loop

 photosDir = 'allImages'
 def getFilesList(self):
        listOfDir = os.listdir(os.path.join(os.getcwd()+photosDir))
        setOfImg = []
        for key,row in enumerate(listOfDir):
            print(os.getcwd()+photosDir+str(row) , 'Image Path')
            setOfImg.append((
                'images['+str(key)+']',(row,open(os.path.join(os.getcwd()+photosDir+'/'+str(row)),'rb'),'image/jpg')
            ))
        print(setOfImg)
        return  setOfImg

Answer 8

In case you have the files from a form and want to forward it to other URL or to API. Here is an example with multiple files and other form data to forward to other URL.

images = request.files.getlist('images')
files = []
for image in images:
    files.append(("images", (image.filename, image.read(), image.content_type)))
r = requests.post(url="http://example.com/post", data={"formdata1": "strvalue", "formdata2": "strvalue2"}, files=files)

Answer 9

Using These methods File will Automatically be closed.

Method 1

with open("file_1.txt", "rb") as f1, open("file_2.txt", "rb") as f2:
    files = [f1, f2]
    response = requests.post('URL', files=files)

But When You're Opening Multiple Files This Can Get Pretty long

Method 2:

files = [open("forms.py", "rb"), open("data.db", "rb")]
response = requests.post('URL', files=files)

# Closing all Files
for file in files: 
    file.close()

Answer 10

The way of files = [("file", (filename, fileobj)), ("file", (filename, fileobj))] in the other answers weren't working for me, but files={"file1": (filename, fileobj), "file2": (filename, fileobj)} did. Just an alternative method in case the above answers don't work