How to cut the output to only gather the filename and get-parameter?

Question

I have multiple files with multiple links that are formatted like this:

<a href="http://example.com/fnord.layername.html?parameter=FOO-_-BAR-_-FNORD" class="poit">
    <img ... />
</a>

Bear in mind that FOO, BAR and FNORD can differ.

Now I want to make sure that all those files have the correct attributes set. For that purpose I want to use ack-grep and cut.

ack-grep -a ?parameter= *.txt

gettiting the output:

fileName.txt
1:<a href="http://example.com/fnord.layername.html?parameter=FOO-_-BAR-_-FNORD" class="poit">

fileName2.txt
5:<a href="http://example.com/fnord.layername.html?parameter=POIT-_-NARF-_-BAZINGA" class="poit">

Yet I am at a total lost about how I could cut the output so that I would only see:

fileName.txt
parameter=FOO-_-BAR-_-FNORD

fileName2.txt
parameter=POIT-_-NARF-_-BAZINGA

Answer 1

I don't have ack-grep utility on my mac so will this awk solution work for you?

awk -v FS=[?\"] '/param/{print FILENAME; print $3 }' INPUT_FILE

Execution:

[jaypal:~/Temp] cat file0
<a href="http://example.com/fnord.layername.html?parameter=FOO-_-BAR-_-FNORD" class="poit">
    <img ... />
</a>

[jaypal:~/Temp] awk -v FS=[?\"] '/param/{print FILENAME; print $3 }' file0
file0
parameter=FOO-_-BAR-_-FNORD

Probable work around for you:

If getting the filename along with parameter values is critical, then you can either run this with a simple for loop or may be you can redirect the output you are getting currently in to a file or pipe it to the following awk one-liner.

Simple for loop:

[jaypal:~/Temp] for i in ./file*.txt; do awk -v FS=[?\"] '/param/{print FILENAME; print $3"\n" }' "$i"; done
./file01.txt
parameter=FOO-_-BAR-_-FNORD

./file02.txt
parameter=POIT-_-NARF-_-BAZINGA

Redirecting your current output to a file and invoking awk on it:

Following example assumes redirection of your current output to a file called temp.file.

[jaypal:~/Temp] cat temp.file 
fileName.txt
1:<a href="http://example.com/fnord.layername.html?parameter=FOO-_-BAR-_-FNORD" class="poit">

fileName2.txt
5:<a href="http://example.com/fnord.layername.html?parameter=POIT-_-NARF-_-BAZINGA" class="poit">

[jaypal:~/Temp] awk -v FS=[?\"] 'NF==1{ print $0 }/param/{ print $3"\n" }' temp.file 
fileName.txt
parameter=FOO-_-BAR-_-FNORD

fileName2.txt
parameter=POIT-_-NARF-_-BAZINGA

Answer 2

You can use the --output EXPR argument to ack-grep to selectively print just part of the line. Keep in mind that ack-grep is a Perl script, the EXPR argument is just any Perl expression; in this case I use parentheses in the pattern match to grab just the part of the line from the ? to the closing " (I also use backslash to quote the regexp metachar '?' which you didn't do, although perhaps as a first character in a regexp that might happen to work).

ack-grep -a --output '$1' '\?(parameter=.*?)"'

Note the last ? in the regexp makes the .* non-greedy, so that it matches up to the first " and not the last one on the line.

This generates output lines like 1:parameter=FOO-_-BAR-_-FNORD - you can suppress the leading 1: with the -h argument to ack-grep.

Answer 3

I'm not sure I understand, and I'm not familiar with the ack-grep utility, but why don't you try cutting from the first occurrence of ? to the " ?

Something like the following:

For readability reason, I put the html tag in a variable

var='<a href="http://example.com/fnord.layername.html?parameter=FOO-_-BAR-_-FNORD" class="poit">'

cut -d '?' -f 2- <(echo $var ) | cut -d '"' -f 1