1

I have this file over here.

    <ul class="social-icons">
        <li><a href=""><span class="social_linkedin_square"></span></a></li>
    </ul>
    <ul class="other-icons">
        <li><a href=""><span class="other-icon"></span></a></li>
    </ul>

I'm trying to only print all the list items in the file using the global command.
(i.e. lines containing <li>...</li> NOTE: <li> and </li> can be on separate lines)

I tried typing this in the command line:

:g/<li>/.,/<\/li>/p

But for whatever reason it's including </ul> <ul class="other-icons"> rather than only the <li>...</li> lines. Alongside with a range error:

        <li><a href=""><span class="social_linkedin_square"></span></a></li>
    </ul>
    <ul class="social-icons">
        <li><a href=""><span class="another"></span></a></li>
E16: Invalid range

Isn't that weird since this global command matches only lines containing <li> and then the print command runs on each of those lines ending with </li> so it should stop the match at that same line? What am I missing?

UPDATE: The command works correctly only if <li> and </li> are on separate lines which is interesting.

4
  • You sure you gave us the same global command that you ran? Because the one above doesn't look right to me and, indeed, causes an Invalid Range error when I run it.
    – B Layer
    Feb 14 at 6:58
  • This is what I get when I run :g/<li>/.,/<\/li>/p <li><a href=""><span class="social_linkedin_square"></span></a></li> </ul> <ul class="social-icons"> <li><a href=""><span class="another"></span></a></li> E16: Invalid range
    – mkayyash
    Feb 14 at 7:49
  • You should include that information in your question. (Hit the "Edit" link to do so.)
    – B Layer
    Feb 14 at 10:00
  • 1
    @BLayer just edited it
    – mkayyash
    Feb 14 at 18:47
2

When you have / characters in your pattern is is better to use another separator, for instance | (see :help :g).

I am guessing that too many separators are confusing the parsing engine.

The following command will do what you want with the example you gave (the <li> elements are all on a single line)

:g|<li>.*</li>|p

Here I am using | as a separator and .* captures everything up to </li>.

6
  • 1
    I think the original intent was « for every line with the start of a list item, print from that line to the line with the closing list item »
    – D. Ben Knoble
    Feb 14 at 21:21
  • Thank you for the answer. So this works assuming <li>and </li> are on the same line which is not always the case. But to be honest, I'm just mostly puzzled at why the specific command that I ran :g/<li>/.,/<\/li>/p would not work correctly.
    – mkayyash
    Feb 14 at 21:40
  • @mkayyash I edited my answer with a guess :-))
    – guntbert
    Feb 14 at 21:47
  • Appreciate trying! I replaced <li> with liopen and </li> with liclose in the buffers. Then I ran :g/liopen/.,/liclose/p and got the same error so probably not a separator issue.
    – mkayyash
    Feb 14 at 21:51
  • Actually if liopen and liclose are on separate lines. Everything works correctly.
    – mkayyash
    Feb 14 at 22:08
2

Reading through the documentation I found the answer. This is not related to the global command but mainly due to cmdline-ranges operating on the NEXT line matching the pattern rather than same or next line which I originally thought.

:h cmdline-ranges
    /{pattern}[/]   the next line where {pattern} matches     *:/*

So running :g/<li>/.,/<\/li>/p is akin to running :.,/<\/li>/p on each line containing <li>. And since /pattern/ matches starting from the NEXT line the original expression was basically unable to find a closing </li> on the last match which causes the range error.

2
  • @guntbert edited to add more details. Let me know if it's missing anything else.
    – mkayyash
    Feb 15 at 22:42
  • Looks good, thanks for your contribution.
    – guntbert
    Feb 17 at 11:16

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