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How to get part of string in awk from match

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# 1  
Old 05-05-2010
How to get part of string in awk from match
Hi,

Im an awk noob and I am having trouble trying to get matches.

Here is my script:

Code:
#!/bin/gawk -f
BEGIN {}
$0 ~ /<a href=".*">.*<\/a>/{print}

Ideally I want to be able to get the actual link and print it. In PHP you can do preg_replace and get the match you want by using \\1 where 1 is the occurance of the match.

I have no idea how to do this in AWK.
# 2  
Old 05-05-2010
Perl to the rescue. If you want the complete line:
Code:
perl -lne 'print if /<a href=".*?">/;' file.html

If you want just the URI:
Code:
perl -lne 'print $1 if /<a href="(.*?)">/;' file.html

If you want just the link text:
Code:
perl -lne 'print $1 if /<a href=".*?">(.*?)<\/a>/;' file.html

# 3  
Old 05-05-2010
Ah, your use of brackets help me devise it in awk:

Code:
$0 ~ /<a href=".*">(.*)<\/a>/ {print gensub(/<a href=\"(.*)\">.*<\/a>/, "\\1", 1)}

Now I just need to find out how it can capture multiple links that are on the same line...?
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