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Asked: 2024-09-07 18:46:35 +0800 CST

help to grep a string from a site

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I'm trying to get a spacific changing string form a site. I used the commend: wget -qO - www.factorio.com/download | grep '/get-download the output is:

    wget -qO - www.factorio.com/download | grep '/get-download'
              <p>If you are running a server, you can use <a href="https://factorio.com/get-download/stable/headless/linux64">this link</a> to always download the latest stable version.
  <a href="/get-download/1.1.110/headless/linux64"
  <a href="/get-download/1.1.110/demo/win64-manual"
  <a href="/get-download/1.1.110/demo/win64"
  <a href="/get-download/1.1.110/demo/osx"
  <a href="/get-download/1.1.110/demo/linux64"

I need to get the 1.1.110 from the line:

<a href="/get-download/1.1.110/headless/linux64"

can be longer and can be shorter. but the /get-download/XXXXX/headless/linux64 is fixed.

command-line

2 Answers

  1. Best Answer

    grep is not the correct tool for extracting information from HTML/XML, but it's sufficient in your case.

    wget -qO - www.factorio.com/download | grep -oP '(?<=/get-download/)[0-9.]+' | uniq

    This matches texts that start with /get-download/ and then contain something somewhat looking like a version number based on numbers and dots. uniq is required, because multiple version lines will be returned. The -o flag lets grep return only matches and <?= will throw away the download text after matching.

    This is the result: 1.1.110

    • 4

  2. Using any version of sed:

    $ wget -qO - www.factorio.com/download |
     sed -n 's:.*/get-download/\([0-9.]*\)/headless/.*:\1:p'
1.1.110

    or any version of grep and cut:

    $ wget -qO - www.factorio.com/download |
    grep '/get-download/[0-9.]*/headless/' | cut -d'/' -f3
1.1.110

    or GNU awk for the 3rd arg to match():

    $ wget -qO - www.factorio.com/download |
    awk 'match($0, "/get-download/([0-9.]+)/headless/", a) { print a[1]; exit }'
1.1.110
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