extract only the substring after double quotes - grep

Question

I have a file which is as below.

<a href="http://firstlink.com" title="title1">
<a href="http://secondlink.com" title="title2">
<a href="http://thirdlink.com" title="title3">
<a href="http://fourthlink.com" title="title4">

I am trying to extract only the URLs from the above file. I am using the below command.

grep -o '\".*\"' new.txt

However, the above command gives me the output as,

"http://firstlink.com" title="title1">
"http://secondlink.com" title="title2">
"http://thirdlink.com" title="title3">
"http://foruthlink.com" title="title4">

I am trying to extract only the URLs without the "". So, my expected output is,

http://firstlink.com
http://secondlink.com
http://thirdlink.com
http://fourthlink.com

How should I change the grep command? Or is it possible to do it in perl, awk or sed command?

Answer 1

You could use awk.

awk -F\" '{print $2}' filename

would produce the desired output.

Using sed:

sed 's/[^"]*"\([^"]*\).*/\1/' filename

Using grep:

grep -oP '[^"]*"\K[^"]*' filename

Answer 2

regexp, stream editors and interpreters are overkill here.
Use the old good cut :

cut -d \" -f 2 < filename

Answer 3

sed 's/.*"\(http.*\)" .*/\1/' filename

Answer 4

This is more portable, since some of the other answers depend on href being the first element

grep -o href.*\" file.txt | cut -d \" -f 2