hey mike-the sample code was very useful. have 2 questions
when i use what you wrote which is listed below i get told
unboundlocalerror: local variable 'product' referenced before
assignment. if i however chnage row to incident in "for incident in
bs('tr'):" i then get mytuples printed out nicely but once again get a
long list of
[('pizza;','pizza hut;', '3.94;')]
[('pizza;','pizza hut;', '3.94;')]
for row in bs('tr'):
data=[]
for incident in row('h2', {'id' : 'dealName'}):
productlist = []
for oText in incident.fetchText( oRE):
productlist.append(oText.strip() + ';')
product = ''.join(productlist)
for incident in row('a', {'name' : 'D0L3'}):
storelist = []
for oText in incident.fetchText( oRE):
storelist.append(oText.strip() + ';')
store = ''.join(storelist)
tuple = (product, store, price)
data.append(tuple)
print data
> [EMAIL PROTECTED] writes:
> > hey kent thanks for your help.
> >
> > so i ended up using a loop but find that i end up getting the same set
> > of results every time. the code is here:
> >
> > for incident in bs('tr'):
> > data2 = []
> > for incident in bs('h2', {'id' : 'dealName'}):
> > product2 = ""
> > for oText in incident.fetchText( oRE):
> > product2 += oText.strip() + ';'
> >
> >
> >
> > for incident in bs('a', {'name' : 'D0L3'}):
> > store2 = ""
> > for oText in incident.fetchText( oRE):
> > store2 += oText.strip() + ';'
> >
> >
> > for incident in bs('a', {'class' : 'nojs'}):
> > price2 = ""
> > for oText in incident.fetchText( oRE):
> > price2 += oText.strip() + ';'
> >
> >
> > tuple2 = (product2, store2, price2)
> > data2.append(tuple2)
> > print data2
>
> Two things here that are bad in general:
> 1) Doing string catenations to build strings. This is slow in
> Python. Build lists of strings and join them, as below.
>
> 2) Using incident as the index variable for all four loops. This is
> very confusing, and certainly part of your problem.
>
> > and i end up getting the following instead of unique results
> >
> > pizza, pizzahut, 3.94
> > pizza, pizzahut, 3.94
> > pizza, pizzahut, 3.94
> > pizza, pizzahut, 3.94
>
> Right. The outer loop doesn't do anything to change what the inner
> loops search, so they do the same thing every time through the outer
> loop. You want them to search the row returned by the outer loop each
> time.
>
> for row in bs('tr'):
> data2 = []
> for incident in row('h2', {'id' :'dealName'}):
> product2list = []
> for oText in incident.fetchText(oRE):
> product2list.append(OText.strip() + ';')
> product2 = ''.join(product2list)
> # etc.
>
> <mike
> --
> Mike Meyer <[EMAIL PROTECTED]>
> http://www.mired.org/home/mwm/
> Independent WWW/Perforce/FreeBSD/Unix consultant, email for more information.
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