Activity log for bug #254901

Date Who What changed Old value New value Message
2008-08-05 09:26:15 Markus Korn bug added bug
2008-08-06 18:53:20 Leonard Richardson launchpadlib: status New Confirmed
2008-08-06 18:53:20 Leonard Richardson launchpadlib: statusexplanation It's not really practical to monitor a list for changes; we'd have to wrap it in a proxy object that knew about the entire Python list interface. We could prevent the problem by deserializing tags into a tuple rather than a list. Then you'd know you had to replace the tuple with another one.
2009-04-20 17:59:00 Leonard Richardson launchpadlib: importance Undecided Low
2009-04-20 17:59:00 Leonard Richardson launchpadlib: status Confirmed Triaged
2009-08-20 20:42:34 Joey Stanford tags oem-services
2011-03-04 20:56:16 Martin Pool description Other then documented in the reference (http://people.ubuntu.com/~flacoste/launchpad-api-doc.html#bug) bug.tags is not a whitespace separated string but a list. It would be nice if launchpadlib could support the related list operations properly, for example: In [38]: b = launchpad.bugs[253799] In [39]: b.tags Out[39]: [u'bar', u'foo'] In [40]: b.tags.append("far") n [41]: b._dirty_attributes Out[41]: {} This only works for changing the complete object: In [42]: b.tags = ["bood"] In [43]: b._dirty_attributes Out[43]: {'tags': ['bood']} Markus Other then documented in the reference (http://people.ubuntu.com/~flacoste/launchpad-api-doc.html#bug) bug.tags is not a whitespace separated string but a list. It would be nice if launchpadlib could support the related list operations properly, for example: In [38]: b = launchpad.bugs[253799] In [39]: b.tags Out[39]: [u'bar', u'foo'] In [40]: b.tags.append("far") n [41]: b._dirty_attributes Out[41]: {} This only works for changing the complete object: In [42]: b.tags = ["bood"] In [43]: b._dirty_attributes Out[43]: {'tags': ['bood']} Markus --- Workaround: rather than appending to the list, set the attribute to a new list that has the value you want: Not: b.tags.append('foo') but rather: b.tags = b.tags + ['foo']
2012-05-22 18:09:35 Bryce Harrington removed subscriber Bryce Harrington