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I want to get the latets version number from this website: https://www.mailpiler.org/wiki/download

so I do a curl to get the content and then just use a pipe to search for the string using grep curl https://www.mailpiler.org/wiki/download | grep piler-

but this way I get the whole line like this: <a href="https://bitbucket.org/jsuto/piler/downloads/piler-1.3.12.tar.gz" class="urlextern" title="https://bitbucket.org/jsuto/piler/downloads/piler-1.3.12.tar.gz" rel="ugc nofollow">piler-1.3.12</a> (~3.3 <abbr title="Megabyte">MB</abbr>), 2022.05.28.

now.. how can I remove everythign but the version number (1.3.12)?

uSlackr
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helpmenicely
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1 Answers1

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With awk you can use command like this:

curl -k -s  https://www.mailpiler.org/wiki/download |awk -F'[-".]' '/piler-/ {OFS=".";print $4,$5,$6}'

you do not need grep because awk can act on similar way.

Romeo Ninov
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