12

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?

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4 Answers 4

Reset to default
16

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
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2
  • 2
    The \K is explained here: stackoverflow.com/a/33573989/318765 Commented Jul 28, 2019 at 23:01
  • The grep command returns too much. Eg. also title=. Commented Sep 14, 2019 at 8:44
12

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

cut -d \" -f 2 < filename
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1
  • I agree, this is exactly what cut was designed for. Commented Feb 16, 2014 at 15:06
2 
sed 's/.*"\(http.*\)" .*/\1/' filename
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1

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
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