7

I have this code:

import requests

r = requests.get('https://www.instagram.com/p/CJDxE7Yp5Oj/?__a=1')
data = r.json()['graphql']['shortcode_media']

Why do I get an error like this?

C:\ProgramData\Anaconda3\envs\test\python.exe C:/Users/Solba/PycharmProjects/test/main.py  
Traceback (most recent call last):  
  File "C:/Users/Solba/PycharmProjects/test/main.py", line 4, in <module>  
    data = r.json()  
  File "C:\ProgramData\Anaconda3\envs\test\lib\site-packages\requests\models.py", line 900, in json  
    return complexjson.loads(self.text, **kwargs)  
  File "C:\ProgramData\Anaconda3\envs\test\lib\json\__init__.py", line 357, in loads  
    return _default_decoder.decode(s)  
  File "C:\ProgramData\Anaconda3\envs\test\lib\json\decoder.py", line 337, in decode  
    obj, end = self.raw_decode(s, idx=_w(s, 0).end())  
  File "C:\ProgramData\Anaconda3\envs\test\lib\json\decoder.py", line 355, in raw_decode  
    raise JSONDecodeError("Expecting value", s, err.value) from None  
json.decoder.JSONDecodeError: Expecting value: line 1 column 1 (char 0)  
  
Process finished with exit code 1  
Karl Knechtel
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Andre Solbach
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    Does this answer your question? [JSONDecodeError: Expecting value: line 1 column 1 (char 0)](https://stackoverflow.com/questions/16573332/jsondecodeerror-expecting-value-line-1-column-1-char-0) – Ulrich Eckhardt Dec 21 '20 at 17:04

2 Answers2

7

r.json() expects a JSON string to be returned by the API. The API should explicitly say it is responding with JSON through response headers.

In this case, the URL you are requesting is either not responding with a proper JSON or not explicitly saying it is responding with a JSON.

You can first check the response sent by the URL by:

data = r.text
print(data)

If the response can be treated as a JSON string, then you can process it with:

import json
data = json.loads(r.text)

Note: You can also check the content-type and Accept headers to ensure the request and response are in the required datatype

veri_pudicha_coder
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0

The reason is because the response is not returning JSON, but instead a whole HTML page. Try r.text instead of r.json()..., and then do whatever you want from there.

If you are not sure the type of content it returns:

h = requests.head('https://www.instagram.com/p/CJDxE7Yp5Oj/?__a=1')
header = h.headers
contentType = header.get('content-type')
print(contentType)

Based on your URL, it returns text/html.

Alternatively, you can try to add a User-Agent in your request - this is to emulate the request to make it look like it comes from a browser, and not a script.

headers = {
    'User-Agent': 'Mozilla/5.0 (Windows NT 10.0; Win64; x64) Chrome/46.0.2490.80'
}

r = requests.get('https://www.instagram.com/p/CJDxE7Yp5Oj/?__a=1', headers=headers)
data = r.json()
blessthefry
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  • If you go on the url: https://www.instagram.com/p/CJDxE7Yp5Oj/?__a=1 You see its already written in json, how can I make it repsonse with that? – Andre Solbach Dec 21 '20 at 17:27