SyntaxError: 'return' outside function
skip at pobox.com
skip at pobox.com
Wed Jan 31 16:49:13 EST 2007
Melih> Has anyone seen this error before and been able to solve it? I
Melih> can't seem to find anything that leads to a solution.
Your code is incorrectly indented. Try:
def legiturl(self, url):
# this breaks down the url into 6 components to make sure it's "legit"
t = urlparse.urlparse(url)
if t[0] != 'http':
return ""
# remove URL fragments, but not URL
if len(t[5]) > 0:
url = urlparse.urlunparse((t[0],t[1],t[2],"","",""))
t = urlparse.urlparse(url)
# stupid parser sometimes leaves frag in path
x = find(t[2], '#')
if x >= 0:
return ""
instead. Note also the lack of a return at the end of the function.
Skip
More information about the Python-list
mailing list