On Feb 1, 3:34 am, asit <lipu...@gmail.com> wrote:
> I hv been developing a link scanner. Here the objective is to
> recursively scan a particular web site.
>
> During this, my script methttp://images.google.co.in/imghp?hl=en&tab=wi
> and passed it to the scan function, whose body is like this..
>
> def scan(site):
> log=open(logfile,'a')
> log.write(site + "\n")
> site = "http://"; + site.lower()
> try:
> site_data = urllib.urlopen(site)
> parser = MyParser()
> parser.parse(site_data.read())
> except(IOError),msg:
> print "Error in connecting site ", site
> print msg
> links = parser.get_hyperlinks()
> for l in links:
> log.write(l + "\n")
>
> But it throws a weird exception like this...
>
> Traceback (most recent call last):
> File "I:\Python26\linkscan1.py", line 104, in <module>
> main()
> File "I:\Python26\linkscan1.py", line 95, in main
> scan(lk)
> File "I:\Python26\linkscan1.py", line 65, in scan
> site_data = urllib.urlopen(site)
> File "I:\Python26\lib\urllib.py", line 87, in urlopen
> return opener.open(url)
> File "I:\Python26\lib\urllib.py", line 203, in open
> return getattr(self, name)(url)
> File "I:\Python26\lib\urllib.py", line 327, in open_http
> h = httplib.HTTP(host)
> File "I:\Python26\lib\httplib.py", line 984, in __init__
> self._setup(self._connection_class(host, port, strict))
> File "I:\Python26\lib\httplib.py", line 656, in __init__
> self._set_hostport(host, port)
> File "I:\Python26\lib\httplib.py", line 668, in _set_hostport
> raise InvalidURL("nonnumeric port: '%s'" % host[i+1:])
> httplib.InvalidURL: nonnumeric port: ''
>
> How can i handle this ???
"Handle" as in how do I catch that exception? The exception's name is
give in the error message. Look at the last line. To catch that
exception, you can do this:
import httplib #that's the module the exception lives in
#as indicated in error message
try:
....
....
....
except httplib.InvalidURL, e:
print e
print "Bad url"
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