How can I make post file requests using requests under python?

Question

I have made a post requests with multipart/form-data; boundary=a1c2469c-2a1f-48e6-8f1d-311f8650c855 And I get this

POST /api/feed/upload-resource HTTP/1.1
App-Id: 15762288
Version-Name: 3.26.0.451
User-Agent: Right-Android/3.26.0.451
Content-Type: multipart/form-data; boundary=a1c2469c-2a1f-48e6-8f1d-311f8650c855
Content-Length: 75412
Connection: Keep-Alive
Accept-Encoding: gzip

--a1c2469c-2a1f-48e6-8f1d-311f8650c855
Content-Disposition: form-data; name="h"
Content-Length: 4

1080
--a1c2469c-2a1f-48e6-8f1d-311f8650c855
Content-Disposition: form-data; name="w"
Content-Length: 4

1080
--a1c2469c-2a1f-48e6-8f1d-311f8650c855
Content-Disposition: form-data; name="image"; filename="/storage/emulated/0/Android/data/com.xiaoyu.rightone/tiny/tiny-738-2018-08-03-15-32-53.jpg"
Content-Type: text/plain
Content-Length: 74910

If I want to simulate this request using Python requests package How can I do that.

I have checked the document, and I have tried use file parameter like this

with open(path, 'rb') as f:
    files = {
        "h": "1495",
        "w": "840",
        "filename": ("image", f.read()),
    }
    r = requests.post(url, files=files)

However in my situation, I always get an error like upload_resource_empty.

zeynel · Accepted Answer · 2018-08-03 10:10:11Z

2

This should work:

import requests

files = {
    'image': ('file_name.jpg', open('file.jpg', 'rb'), 'text/plain'),
    'w': (None, '123'),
    'h': (None, '222')
}

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

The tuples in the dictionary have the following format: ('filename', fileobj, 'content_type', custom_headers)

answered Aug 3, 2018 at 10:10

zeynel

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1 Comment

einverne Over a year ago

I have figured out that server side handle the w and h in post body. So I send w and h in post data, and solved. Thank you for your information.

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