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
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
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