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Unwanted mapping using? with grep

I am writing a bash script that parses an html file and I want to get the contents of each <tr>...</tr> . Therefore, my command looks like this:

 $ tr -d \\012 < price.html | grep -oE '<tr>.*?</tr>'

But grep seems to give me the result:

 $ tr -d \\012 < price.html | grep -oE '<tr>.*</tr>'

How can I do .* Not greedy?

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bash regex grep
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4 answers

If you have GNU Grep , you can use -P to make the match inanimate:

 $ tr -d \\012 < price.html | grep -Po '<tr>.*?</tr>'

The -P allows you to execute the Perl Compliment (PCRE) regular expression, which is necessary for non-greedy matching with ? because Basic Regular Expression (BRE) and Extended Regular Expression (ERE) do not support it.

If you use -P , you can also use look around to avoid printing tags in the match like this:

 $ tr -d \\012 < price.html | grep -Po '(?<=<tr>).*?(?=</tr>)'

If you don't have GNU Grep and the HTML is well-formed, you can simply do:

 $ tr -d \\012 < price.html | grep -o '<tr>[^<]*</tr>'

Note. The above example will not work with nested tags inside <tr> .

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Unwanted matching is not part of the Extended Regular Expression syntax supported by grep -E . Use grep -P instead if you have it, or switch to Perl / Python / Ruby / what you have. (Oh and pcregrep .)

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.*? is a regular expression of Perl. Change grep to

 grep -oP '<tr>.*?</tr>'
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Try perl-style-regexp

 $ grep -Po '<tr>.*?</tr>' input <tr>stuff</tr> <tr>more stuff</tr>
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