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Display URL as part of HTTP error messages

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It seems convenient to include the URL in the error message in case you
get an unexpected error.

E.g.:

    In [1]: import requests

    In [2]: resp = requests.get('http://www.google.com/eofdfdfdfdfd')

    In [3]: resp
    Out[3]: <Response [404]>

    In [4]: resp.raise_for_status()
    ---------------------------------------------------------------------------
    HTTPError                                 Traceback (most recent call last)
    <ipython-input-4-00e7077cfb5b> in <module>()
    ----> 1 resp.raise_for_status()

    /Users/marca/dev/git-repos/requests/requests/models.py in raise_for_status(self)
        835
        836         if http_error_msg:
    --> 837             raise HTTPError(http_error_msg, response=self)
        838
        839     def close(self):

    HTTPError: 404 Client Error: Not Found for url: http://www.google.com/eofdfdfdfdfd
This commit is contained in:
Marc Abramowitz
2015-06-22 13:22:59 -07:00
parent 9bbab338fd
commit e3bdec5934
1 changed files with 2 additions and 2 deletions
Download Patch File Download Diff File Expand all files Collapse all files
requests/models.py
+2 -2
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@@ -828,10 +828,10 @@ class Response(object):
http_error_msg = ''
if 400 <= self.status_code < 500:
http_error_msg = '%s Client Error: %s' % (self.status_code, self.reason)
http_error_msg = '%s Client Error: %s for url: %s' % (self.status_code, self.reason, self.url)
elif 500 <= self.status_code < 600:
http_error_msg = '%s Server Error: %s' % (self.status_code, self.reason)
http_error_msg = '%s Server Error: %s for url: %s' % (self.status_code, self.reason, self.url)
if http_error_msg:
raise HTTPError(http_error_msg, response=self)