> Gerard Flanagan wrote:
> > 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, "[EMAIL PROTECTED]" % routeid)
> > #changed line
> > if existing_route:
> > route = existing_route[0]
> > existing_equip = findall(route, "[EMAIL PROTECTED]'%s']" % equipid)
> > if existing_equip:
> > [...]
> >
> > ie. don't quote the route id since it's numeric.
kepioo wrote:
> thanks a lot for the code.
>
> It was not working the first time (do not recognize item and
> existing_time --
Apologies, I ran the code from PythonWin which remembers names that
were previously declared though deleted - should have run it as a
script.
> i changed item by r[-1] and existing_time by
> existing_equip).
>
'item' was wrong but not the other two. (I'm assuming your data is
regular - ie. all the records have the same number of fields)
change the for loop to the following:
8<------------------------------------------------------
for routeid, equipid, timeid, data in records:
route, equip, time = None, None, None
existing_route = findall(results, "[EMAIL PROTECTED]" % routeid)
if existing_route:
route = existing_route[0]
existing_equip = findall(route, "[EMAIL PROTECTED]" % equipid)
if existing_equip:
equip = existing_equip[0]
existing_time = findall(equip, "[EMAIL PROTECTED]" % timeid)
if existing_time:
time = existing_time[0]
route = route or SubElement(results, 'route', id=routeid)
equip = equip or SubElement(route, 'equip', id=equipid)
time = time or SubElement(equip, 'time', id=timeid)
dataitem = SubElement(time,'data')
dataitem.text = data
8<------------------------------------------------------
> however, it is not producing the result i expected, as in it doesn't
> group by same category the elements, it creates a new block of xml
>
[...]
the changes above should give you what you want - remember, as I wrote
in the previous post, it should be:
"[EMAIL PROTECTED]"
not
"[EMAIL PROTECTED]'%s']"
ie. no single quotes needed.
With the above amended code I get:
<results>
<route id="23">
<equip id="jr2">
<time id="3pm">
<data>my first value</data>
<data>my third value</data>
</time>
</equip>
<equip id="jr1">
<time id="3pm">
<data>my second value</data>
</time>
</equip>
</route>
<route id="24">
<equip id="jr2">
<time id="3pm">
<data>my fourth value</data>
</time>
</equip>
</route>
<route id="25">
<equip id="jr2">
<time id="3pm">
<data>my fifth value</data>
</time>
<time id="4pm">
<data>my sixth value</data>
</time>
</equip>
</route>
</results>
------------------------------------
all the best
Gerard
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