grep - Spider a Website and Return URLs Only

Question

I'm looking for a way to pseudo-spider a website. The key is that I don't actually want the content, but rather a simple list of URIs. I can get reasonably close to this idea with Wget using the --spider option, but when piping that output through a grep, I can't seem to find the right magic to make it work:

wget --spider --force-html -r -l1 http://somesite.com | grep 'Saving to:'

The grep filter seems to have absolutely no affect on the wget output. Have I got something wrong or is there another tool I should try that's more geared towards providing this kind of limited result set?

UPDATE

So I just found out offline that, by default, wget writes to stderr. I missed that in the man pages (in fact, I still haven't found it if it's in there). Once I piped the return to stdout, I got closer to what I need:

wget --spider --force-html -r -l1 http://somesite.com 2>&1 | grep 'Saving to:'

I'd still be interested in other/better means for doing this kind of thing, if any exist.

Answer 1

The absolute last thing I want to do is download and parse all of the content myself (i.e. create my own spider). Once I learned that Wget writes to stderr by default, I was able to redirect it to stdout and filter the output appropriately.

wget --spider --force-html -r -l2 $url 2>&1 
  | grep '^--' | awk '{ print $3 }' 
  | grep -v '.(css|js|png|gif|jpg)$' 
  > urls.m3u

This gives me a list of the content resource (resources that aren't images, CSS or JS source files) URIs that are spidered. From there, I can send the URIs off to a third party tool for processing to meet my needs.

The output still needs to be streamlined slightly (it produces duplicates as it's shown above), but it's almost there and I haven't had to do any parsing myself.