Urllib2 in python: unable to get page, how to go to next page in this case?

Question

Python

I get the page in python like this:

for x in range(1, 10):
   response = urllib2.urlopen("http://site.com/id="+str(x))
   page = response.read()

Sometimes there are such problems that the page cannot be retrieved. How can I go to the next page in this case?
That is, let's say we are trying to get the site.com/id=3 page, but it does not respond. How do I navigate to site.com/id=4?

Answer 1

If it throws an exception - catch it with try\except , then the program will not be interrupted.

Answer 2

You have 2 options:
1. Before sending a get request, send a head request in order to check what code will be returned (in this case, no exception is thrown).
2. Catch the get request dispatch exception as @SolidlSnake said with try/except blocks.
Pretty clear examples of both of these options are given here:
stackoverflow.com/a/16778473