github.com_openSUSE_osc

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Marcus Huewe 81d1985bc5 Improve error message in case of an URLError

The old code does not print any information about the host, for
which the access failed, in case of an URLError. In order to fix
this, add information about the host (and port) to the URLError
instance in core.http_request and use this information in the
babysitter to print out a more detailed error message (which includes
the host (and port)).
For now, we simply add a "private" "_osc_host_port" attribute to
the URLError instance (this way we avoid potential name clashes (due
to the "_osc" prefix) and could come up with a different/more clever
way in the future (due to its privateness)).

Fixes: #954 ("Better diagnostic for domain name issues")

2021-09-30 14:09:08 +02:00

util

Use os.getcwdb() instead of os.getcwd().encode() in util.cpio.CpioRead

2020-11-22 17:39:54 +01:00

__init__.py

replace urlgrabber to enable python3 compatibility

2018-10-19 09:31:37 +02:00

.gitignore

convert svn:ignore to gitignore

2009-12-03 19:19:53 +01:00

babysitter.py

Improve error message in case of an URLError