Writing API get request output to file only writes the status code

Question

I'm trying to write the output of an API 'get' request to a file. I know the function works as the output of the request is returned successfully and in full in shell.

The function creates the file, but the file only includes the status code for the get request - can anyone help me understand why? Still pretty new to Python and very new to calling APIs...

def user_info(token):
  heads = {"Authorization": "bearer " + token, "Content-type": "application/json"}
  url = "xxxxx"
  request = req.get(url, headers=heads)
  with open('userdata.txt', 'w') as f:
      f.write(str(request))
  if request.status_code == 200:
      return request.json()
  else:
      return "User Info FAILED! Failure code: {}, error: {}".format(request.status_code,request.json()['error'])

The file just contains:

Thanks!

well what do you want to put in the file? if you want to put the json response then f.write(request.json()) — Chris Doyle
– Chris Doyle, Commented Feb 10, 2021 at 17:03
aha... f.write(str(request.json())) works, but if I don't specify it should be returned as a string is errors with "TypeError: write() argument must be str, not dict" — Steve
– Steve, Commented Feb 10, 2021 at 17:04

Chris Doyle · Accepted Answer · 2021-02-10 17:15:52Z

1

A Response object in the requests module when called as a string will return its status code as detailed in the Response python code

    def __repr__(self):
        return '<Response [%s]>' % (self.status_code)

If you want to write the json from the response as a string then you can use the built in json module and use its dumps method to dump it as a string not a dict.

import requests
import json

with open("myfile.txt", "w") as output:
    resp = requests.get("https://jsonplaceholder.typicode.com/todos/1")
    output.write(json.dumps(resp.json()))

answered Feb 10, 2021 at 17:15

Chris Doyle

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Writing API get request output to file only writes the status code

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