Non greedy matching using ? with grep

Question

I'm writing a bash script which analyses a html file and I want to get the content of each single <tr>...</tr>. So my command looks like:

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

But it seems that grep gives me the result of:

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

How can I make .* non-greedy?

Answer 1

If you have GNU Grep you can use -P to make the match non-greedy:

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

The -P option enables Perl Compliant Regular Expression (PCRE) which is needed for non-greedy matching with ? as Basic Regular Expression (BRE) and Extended Regular Expression (ERE) do not support it.

If you are using -P you could also use look arounds to avoid printing the tags in the match like so:

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

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If you don't have GNU grep and the HTML is well formed you could just do:

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

Note: The above example won't work with nested tags within <tr>.

Answer 2

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

Of course, if you really mean

<tr>[^<>]*</tr>

you should say that instead; then plain old grep will work fine.

You could (tediously) extend the regex to accept nested tags which are not <tr> but of course, it's better to use a proper HTML parser than spend a lot of time rediscovering why regular expressions are not the right tool for this.

Answer 3

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

grep -oP '<tr>.*?</tr>'