simplejson.errors.JSONDecodeError?
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BinbinJin commented
There is an error when I run these codes. The error is simplejson.errors.JSONDecodeError: Expecting value: line 1 column 1 (char 0).
I think the reason is that, in line 79 of url_generator.py, we get the content in HTML form by requests.get(seed). Thus, it can not be transformed into json format by r.json(). Am I right? If so, how should I revise your codes to crawl the data from yahoo answers?
Thank you!