SyntaxError: 'return' outside function

[Steven Bethard]

Melih Onvural wrote:
> Has anyone seen this error before and been able to solve it? I can't
> seem to find anything that leads to a solution. I found this post
> http://zope.org/Collectors/Zope/1809, but can't really understand it.
> I've attached my code below to see if anything looks funny. It happens
> at the very last return at the end. Thanks in advance,
> 
> --melih
> 
> ===========Code========
> 
> 	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 ""

Looks like you're mixing tabs and spaces.  Your "return" statement is 
outside the "legiturl" function.

Steve