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I'm have server on flask

@app.route('/', methods=['POST'])
def main():
    if request.method == 'POST':
        return send_from_directory(directory='./', filename='data.zip')

I send the request with 

import requests
r = requests.post("http://127.0.0.1:5002", data=data)

I'm get

Response [200]

How can i save the r in data.zip on client?

Alexey
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    if server sends .zip file then simply use byte more `f = open('data.zip, 'wb'`)` and `f.writen(r.content)`, `f.close()`. If file is big then you can use `post(..., stream=True)` and use loop to save in chunks. See [streaming requests](https://requests.readthedocs.io/en/master/user/advanced/#streaming-requests) – furas May 22 '20 at 12:38

1 Answers1

2

Use open(), read() and close() with byte mode wb

import requests

r = requests.post("http://127.0.0.1:5002", data=data)

f = open('data.zip', 'wb')
f.write(r.content)
f.close()

Or using with

import requests

r = requests.post("http://127.0.0.1:5002", data=data)

with open('data.zip', 'wb') as f:
    f.write(r.content)

If it is bigger file then you can stream data to read and save in chunks

import requests

r = requests.post("http://127.0.0.1:5002", data=data, stream=True)

with open('data.zip', 'wb') as f:
     for chunk in r.iter_content(chunk_size=8192): 
         f.write(chunk)

Requests doc: Streaming Requests

BTW: similar question on Stackoverflow: Download large file in python with requests

furas
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