Question

Hey i got the following code:

import urllib

source = urllib.urlopen('WEBPAGE').read()
if 'STRING TO LOOK FOR' in source:
    print 'MESSAGE'
else:
    print 'else MESSAGE'

which checks a webpage source code for a specific string, is there a way that only the first lets say 20 lines of the code get searched through for the string? and if the string is not in there i get the else Message?

Answer 1

Change if 'STRING TO LOOK FOR' in source: to

if 'STRING TO LOOK FOR' in '\n'.join(source.split('\n', 20)[:20]):

Split the source into lines and then reassemble the first 20 lines.