xml aggregator
Gerard Flanagan
grflanagan at yahoo.co.uk
Sun Jul 9 12:21:29 EDT 2006
Gerard Flanagan wrote:
> kepioo wrote:
> > Hi all,
> >
> > I am trying to write an xml aggregator, but so far, i've been failing
> > miserably.
> >
> > what i want to do :
> >
> > i have entries, in a list format :[[key1,value],[key2,value],[
> > key3,value]], value]
> >
> > example :
> > [["route","23"],["equip","jr2"],["time","3pm"]],"my first value"]
> > [["route","23"],["equip","jr1"],["time","3pm"]],"my second value"]
> > [["route","23"],["equip","jr2"],["time","3pm"]],"my third value"]
> > [["route","24"],["equip","jr2"],["time","3pm"]],"my fourth value"]
> > [["route","25"],["equip","jr2"],["time","3pm"]],'"my fifth value"]
> >
>
> [snip example data]
>
> >
> >
> > If anyone has an idea of implemetation or any code ( i was trying with
> > ElementTree...
> >
>
> (You should have posted the code you tried)
>
> The code below might help (though you should test it more than I have).
> The 'findall' function comes from here:
>
> http://gflanagan.net/site/python/elementfilter/elementfilter.py
>
> it's not the elementtree one.
>
Sorry, elementfilter.py was a bit broken - fixed now. Use the current
one and change the code I posted to:
[...]
existing_route = findall(results, "route[@id==%s]" % routeid)
#changed line
if existing_route:
route = existing_route[0]
existing_equip = findall(route, "equip[@id=='%s']" % equipid)
if existing_equip:
[...]
ie. don't quote the route id since it's numeric.
Gerard
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