I want to extract all information from a bug (comments, creator's name and email, QA's name and email, etc) and save it to a text file but I'm not able to. I can only access some of the data. I'm using Bugzilla version 4.2.5. Using *print dir(bz.getbug(8658))*, here is what I have access to the following (see attachment).
So long story short, I cannot/don't have access to *description*, *dupe_of* , *assigned_to*/*qa_contact *(can only get email but not the name), *estimated_time*, *actual_time*, *remaining_time*, getting all *comments *(with date, text, author, attachments reference) and *attachments *(file, date, author, description).
But some of these attributes are available as tags on the bug's XML page. So I thought maybe I could just parse the XML page and get all the info from there. Do you have any suggestions? Otherwise I'm stuck because I can't get all the info I need.
Thanks, Ravi
On Tue, Jan 13, 2015 at 4:04 PM, Cole Robinson crobinso@redhat.com wrote:
Sorry I didn't consider the auth issue.
Modern bugzilla doesn't use cookies via the xmlrpc API, instead it uses a token which is API access only. So authenticating via python-bugzilla is not going to give you any way of accessing a regular bugzilla URL with auth. So there's no easy way that I can think of.
What are you trying to achieve exactly? Why do you need the bug XML output? All that information should be available via the python-bugzilla API, so maybe you can achieve what you need in a different way
- Cole
On 01/13/2015 03:56 PM, Ravikumar Patel wrote:
Thanks for the help. Could you help me with that because I am having
login
issues from a couple of stuff I found on the web? Or can you provide me
with
any helpful links?
So far I have looked at these and I keep getting a dummy html page (not related to the bug) instead of the XML page for the bug:
http://stackoverflow.com/questions/189555/how-to-use-python-to-login-to-a-we...
http://stackoverflow.com/questions/11167419/get-xml-data-from-bugzilla-url
On Tue, Jan 13, 2015 at 3:38 PM, Cole Robinson <crobinso@redhat.com mailto:crobinso@redhat.com> wrote:
I guess you can just generate that URL from the bug's ID, then use
standard
python urllib or similar to fetch the web page contents. I don't
think there's
any bugzilla API call that does what you want though - Cole On 01/13/2015 03:35 PM, Ravikumar Patel wrote: > I'll guide you through the steps I've followed to clarify what I
mean. First,
> I queried for all bugs for a specific person. I then got a list of
bugs for
> that person with all of their bug IDs. I clicked on one of the bug
ID and it
> gave me a page with that bug's summary, status, aliases, product,
etc. Once
> you get to the bottom of the page, you have a link to it's XML
view/page. (see
> screenshot below) > > Inline image 1 > > Now I want to get access to this page (using my python script from
previous
> emails): > Inline image 3 > > I want to know if I can access this XML page from a corresponding
bug ID via a
> python script? > > Thanks, > Ravi > > > On Tue, Jan 13, 2015 at 2:25 PM, Cole Robinson <
crobinso@redhat.com mailto:crobinso@redhat.com
> <mailto:crobinso@redhat.com <mailto:crobinso@redhat.com>>> wrote: > > I don't know what you mean by 'xml page' > > - Cole > > On 01/13/2015 02:23 PM, Ravikumar Patel wrote: > > This is a follow up question. How do I now access each bug's
XML page now that
> > I have the bug IDs I needed? > > > > On Tue, Jan 13, 2015 at 1:34 PM, Cole Robinson <
crobinso@redhat.com mailto:crobinso@redhat.com
<mailto:crobinso@redhat.com <mailto:crobinso@redhat.com>> > > <mailto:crobinso@redhat.com <mailto:crobinso@redhat.com> <mailto:crobinso@redhat.com <mailto:crobinso@redhat.com>>>> wrote: > > > > On 01/13/2015 12:59 PM, Ravikumar Patel wrote: > > > Hello, > > > > > > I am trying to get the bug ID of all bugs from my
Bugzilla account. Any
> > > suggestions as to how I am able to do so? Is it
possible that to query for all
> > > projects and then access the bug IDs from each
project? If so, then how?
> > > > > > So far my code looks like this: > > > > > > |bz = bugzilla.Bugzilla(url='
https://bugzilla.mycompany.com/xmlrpc.cgi')
> > > try: > > > bz.login('name@email.com <mailto:name@email.com>
<mailto:name@email.com
<mailto:name@email.com>> <mailto:name@email.com <mailto:
name@email.com>
> <mailto:name@email.com <mailto:name@email.com>>> > > <mailto:name@email.com <mailto:name@email.com> <mailto:
name@email.com
<mailto:name@email.com>> > <mailto:name@email.com <mailto:name@email.com> <mailto:
name@email.com
<mailto:name@email.com>>>>', 'password'); > > > print'Authorization cookie received.' > > > except bugzilla.BugzillaError: > > > print(str(sys.exc_info()[1])) > > > sys.exit(1)| > > > > > > Here is the link to the stackoverflow question page: > > >
http://stackoverflow.com/questions/27869663/how-do-i-query-bugzilla-to-get-a...
> > > > > > I really need help on this. Any suggestions? > > > > bugs =
bz.query(bz.build_query(assigned_to="your-bugzilla-account"))
> > for bug in bugs: > > print bug.id <http://bug.id> <http://bug.id> <
> > > > - Cole > > > > > > > > > >