ok the syntax for next is this: b = n.findNext
LordLaraby wrote: > You wrote: > > i have an > > href which looks like this: > > <td class="all"> > > <a class="btn" name="D1" href="http://www.cnn.com";> > > </a> > > here is my code > > for incident in row('td', {'class':'all'}): > > n = incident.findNextSibling('a', {'class': 'btn'}) > > link = incident.findNextSibling['href'] + "','" > > any idea what i'm doing wrong here with the syntax? thanks in advance > Since you already found the anchor tag with the 'btn' class attribute, > just grab it's href attribute like so: > for incident in row('td', {'class':'all'}): > n = incident.findNextSibling('a', {'class': 'btn'}) > link = n['href'] + "','" > > Works for me... > > LL -- http://mail.python.org/mailman/listinfo/python-list
