How to catch 404 error in urllib.urlretrieve

Question

Background: I am using urllib.urlretrieve, as opposed to any other function in the urllib* modules, because of the hook function support (see reporthook

Answer 1

You should use:

import urllib2

try:
    resp = urllib2.urlopen("http://www.google.com/this-gives-a-404/")
except urllib2.URLError, e:
    if not hasattr(e, "code"):
        raise
    resp = e

print "Gave", resp.code, resp.msg
print "=" * 80
print resp.read(80)

Edit: The rationale here is that unless you expect the exceptional state, it is an exception for it to happen, and you probably didn't even think about it -- so instead of letting your code continue to run while it was unsuccessful, the default behavior is--quite sensibly--to inhibit its execution.