CODEWITHSUNDEEP
jsonpython-3.xdictionary
Oborzil
Oborzil

Reputation: 71

How to avoid JSONDecodeError?

I have a POST request and it returns a JSON. If the user is searching for something what is does not exist, then the JSON is empty. How can I avoid the error message and print something like: 'no results' ?

The error message:

File "/usr/lib/python3.7/json/__init__.py", line 348, in loads
    return _default_decoder.decode(s)   File "/usr/lib/python3.7/json/decoder.py", line 337, in decode
    obj, end = self.raw_decode(s, idx=_w(s, 0).end())   File "/usr/lib/python3.7/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)

The code: // It works well, except if the return JSON is empty //

import requests
import json
from private import payload, criterias


r = requests.session()

url_login = 'https://somesite.com/login.php'

url_search = 'https://somesite.com/data.php?jsons=true'


req1 = r.post(url_login, data=payload)

req2 = r.post(url_search, data=criterias)

data = json.loads(req2.text)


release = 'release_name'

for release in data['results']:
    print(release['release_name']

Upvotes: 0

Views: 696

Answers (1)

Mike Slinn
Mike Slinn

Reputation: 8405

You asked how to avoid JSONDecodeError. The answer is to catch it.

import requests
import json
from private import payload, criterias

r = requests.session()
url_login = 'https://somesite.com/login.php'
url_search = 'https://somesite.com/data.php?jsons=true'

req1 = r.post(url_login, data=payload)
req2 = r.post(url_search, data=criterias)

try:
    data = json.loads(req2.text)
    release = 'release_name'
    for release in data['results']:
        print(release['release_name'])  # your code did not have a close parentheses
except ValueError as e:  # catches subclass JSONDecodeError
    print("No results")
    print(f"req.text={req.text}")
    print(e)

Upvotes: 1

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