Saving response from Requests to file

You can use the response.text to write to a file:

import requests

files = {'f': ('1.pdf', open('1.pdf', 'rb'))}
response = requests.post("https://pdftables.com/api?&format=xlsx-single",files=files)
response.raise_for_status() # ensure we notice bad responses
file = open("resp_text.txt", "w")
file.write(response.text)
file.close()
file = open("resp_content.txt", "w")
file.write(response.text)
file.close()

I believe all the existing answers contain the relevant information, but I would like to summarize.

The response object that is returned by requests get and post operations contains two useful attributes:

Response attributes

  • response.text - Contains str with the response text.
  • response.content - Contains bytes with the raw response content.

You should choose one or other of these attributes depending on the type of response you expect.

  • For text-based responses (html, json, yaml, etc) you would use response.text
  • For binary-based responses (jpg, png, zip, xls, etc) you would use response.content.

Writing response to file

When writing responses to file you need to use the open function with the appropriate file write mode.

  • For text responses you need to use "w" - plain write mode.
  • For binary responses you need to use "wb" - binary write mode.

Examples

Text request and save

# Request the HTML for this web page:
response = requests.get("https://stackoverflow.com/questions/31126596/saving-response-from-requests-to-file")
with open("response.txt", "w") as f:
    f.write(response.text)

Binary request and save

# Request the profile picture of the OP:
response = requests.get("https://i.stack.imgur.com/iysmF.jpg?s=32&g=1")
with open("response.jpg", "wb") as f:
    f.write(response.content)

Answering the original question

The original code should work by using wb and response.content:

import requests

files = {'f': ('1.pdf', open('1.pdf', 'rb'))}
response = requests.post("https://pdftables.com/api?&format=xlsx-single",files=files)
response.raise_for_status() # ensure we notice bad responses
file = open("out.xls", "wb")
file.write(response.content)
file.close()

But I would go further and use the with context manager for open.

import requests

with open('1.pdf', 'rb') as file:
    files = {'f': ('1.pdf', file)}
    response = requests.post("https://pdftables.com/api?&format=xlsx-single",files=files)

response.raise_for_status() # ensure we notice bad responses

with open("out.xls", "wb") as file:
    file.write(response.content)