使用awk sed或grep来解析来自网页源的URL

Question

I am trying to parse the source of a downloaded web-page in order to obtain the link listing. A one-liner would work fine. Here's what I've tried thus far:

This seems to leave out parts of the URL from some of the page names.

$ cat file.html | grep -o -E '\b(([\w-]+://?|domain[.]org)[^\s()<>]+(?:\([\w\d]+\)|([^[:punct:]\s]|/)))'|sort -ut/ -k3

This gets all of the URL's but I do not want to include links that have/are anchor links. Also I want to be able to specify the domain.org/folder/:

$ awk 'BEGIN{
RS="</a>"
IGNORECASE=1
}
{
  for(o=1;o<=NF;o++){
    if ( $o ~ /href/){
      gsub(/.*href=\042/,"",$o)
      gsub(/\042.*/,"",$o)
      print $(o)
    }
  }
}' file.html

Answer 1

If you are only parsing something like < a > tags, you could just match the href attribute like this:

$ cat file.html | grep -o -E 'href="([^"#]+)"' | cut -d'"' -f2 | sort | uniq

That will ignore the anchor and also guarantee that you have uniques. This does assume that the page has well-formed (X)HTML, but you could pass it through Tidy first.

Answer 2

lynx -dump http://www.ibm.com

And look for the string 'References' in the output. Post-process with sed if you need to.

Using a different tool sometimes makes the job simpler. Once in a while, a different tool makes the job dead simple. This is one of those times.