[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/
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