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_ofassigned_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-webpage-and-retrieve-cookies-for-later-usage
> 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-all-the-bug-ids-using-python-and-xml-rpc
>     >     >     >
>     >     >     > 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> <http://bug.id>
>     >     >
>     >     >     - Cole
>     >     >
>     >     >
>     >     >
>     >     >
>     >
>     >
>
>