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Only decode and return browser content in the successful case #162

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Only decode and return browser content in the successful case #162

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9 changes: 7 additions & 2 deletions openqa_review/browser.py
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Expand Up @@ -144,8 +144,13 @@ def _get(self, url, as_json=False): # pragma: no cover
log.warn(msg)
raise DownloadError(msg)

content = r.json() if as_json else r.content.decode("utf8")
return content
try:
content = r.json() if as_json else r.content.decode("utf8")
return content
except json.decoder.JSONDecodeError as e:
msg = "Failed to decode {}: {}".format(r.content, str(e))
log.warn(msg)
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Before we escalate the message with "warning" level I would like to understand what can be the reason for such cases encountered. Should we ignore such issues? Then make it less than "warn". Should we handle such issues? Then "warn" is not high enough.

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I don't know the reason. All i know is that we get invalid JSON, which I'm surfacing here, that the traceback doesn't contain. And I'd consider that an error.

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It would be nice if the URL could be included in the error message.

raise DownloadError(msg)

def json_rpc_get(self, url, method, params, cache=True):
"""Execute JSON RPC GET request."""
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