Joe <[EMAIL PROTECTED]> writes: > start = s.find('<a href="somefile') + len('<a > href="somefile') > stop = s.find('">Save File</a></B>', > start) fileName = s[start:stop] > and then construct the url with the filename to download the image > which works fine as cause every image has the Save File link and I can > count number of images easy the problem is when there is more than image I > try using while loop downlaod files, wirks fine for the first one but > always matches the same, how can count and thell the look to skip the fist > one if it has been downloaded and go to next one, and if next one is > downloaded go to next one, and so on.
To answer your question, use the first optional argument to find in both invocations of find: stop = 0 while end >= 0: start = s.find('<a href="somefile', stop) + len('<a href="somefile') stop = s.find('">Save File</a></B>', start) fileName = s[start:stop] Now, to give you some advice: don't do this by hand, use an HTML parsing library. The code above is incredibly fragile, and will break on any number of minor variations in the input text. Using a real parser not only avoids all those problems, it makes your code shorter. I like BeautifulSoup: soup = BeautifulSoup(s) for anchor in soup.fetch('a'): fileName = anchor['href'] to get all the hrefs. If you only want the ones that have "Save File" in the link text, you'd do: soup = BeautifulSoup(s) for link in soup.fetchText('Save File'): fileName = link.findParent('a')['href'] <mike -- Mike Meyer <[EMAIL PROTECTED]> http://www.mired.org/home/mwm/ Independent WWW/Perforce/FreeBSD/Unix consultant, email for more information. -- http://mail.python.org/mailman/listinfo/python-list