sed command to find .css file inside link

Question

I am using sed to read the .css file name (after "href=") from an html file. The command is follow:

cssFiles=$(echo "$BODY" | sed -rn 's/<link\s.*href=\W(.*.css).*/\1/p')

But, it does not works correctly. Below, sample input, output and expected output is given. Where am I wrong?

Sample input:

<link href="/css/default.css" rel="stylesheet" type="text/css" />
<link rel="stylesheet" type="text/css" href="js/flexslider/flexslider.css">

Sample output:

/css/default.css" rel="stylesheet" type="text/css
js/flexslider/flexslider.css

Expected output:

/css/default.css
js/flexslider/flexslider.css

scopchanov · Accepted Answer · 2018-08-15 15:18:17Z

1

Try this:

cssFiles=$(echo "$BODY" | sed -rn 's/<link\s.*href=\W(.*.css).*/\1/p' | awk -F'=' '{print$2}'  awk -F' ' '{print$1}')

scopchanov

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answered Aug 15, 2018 at 14:16

Kleudy Silva

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