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Asked: 2026-05-14T19:41:03+00:00

Edit: To be clear, please understand that I am not using Regex to parse

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Edit: To be clear, please understand that I am not using Regex to parse the html, that’s crazy talk! I’m simply wanting to clean up a messy string of html so it will parse

Edit #2: I should also point out that the control character I’m using is a special unicode character – it’s not something that would ever be used in a proper tag under any normal circumstances

Suppose I have a string of html that contains a bunch of control characters and I want to remove the control characters from inside tags only, leaving the characters outside the tags alone.

For example

Here the control character is the numeral “1”.

Input

The quick 1<strong>orange</strong> lemming <sp11a1n 1class1='jumpe111r'11>jumps over</span> 1the idle 1frog

Desired Output

The quick 1<strong>orange</strong> lemming <span class='jumper'>jumps over</span> 1the idle 1frog

So far I can match tags which contain the control character but I can’t remove them in one regex. I guess I could perform another regex on my matches, but I’d really like to know if there’s a better way.

My regex

Bear in mind this one only matches tags which contain the control character.

<(([^>])*?`([^>])*?)*?>

Thanks very much for your time and consideration.

Iain Fraser

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1 Answer

  1. Editorial Team

    Regex isn’t the tool for this, but you can use lookbehind and lookahead to match 1 in a tag. Here it is in Java, modified to have finite lookbehind (since Java doesn’t support infinite length lookbehind).

        String s = "123 <o123o></o1o1> <oo 11='11x'> x11 <msg136='I <3 Johnny!11'>";
    System.out.println(
        s.replaceAll("(?<=<[^<>]{0,999})(?=[^<>]+>)1", "")
    ); // prints "123 <o23o></oo> <oo ='x'> x11 <msg136='I <3 Johnny!'>

    There are many cases where this will fail, but it should get you started somewhere.

    See also

    Okay, I’ve “generalized” the problem so that it’s not HTML related. Here’s a snippet of Java that uses regex to remove [aeiou] from portions of a sentence enclosed by < and >, whose usage is reserved only to mark these special portions.

    BEWARE: this regex is absolutely unreadable. But yes, it works. And it uses no lookbehind, too.

    String s = "Wait <whaaat?> does this <really really> work???";
System.out.println(
    s.replaceAll("(?!>)(?:(?=<)|(?=\\G)(?!^))(?:(?:(?![aeiou])(.))|.)", "$1")
); // prints "Wait <wht?> does this <rlly rlly> work???"

    I might try to explain it if there’s interest, but otherwise I’d suggesting using a simple loop like this instead:

    allocate output buffer
set isInside := false
for every character ch in input
   if (ch is openChar)
      isInside := true
   else if (ch is closeChar)
      isInside := false
   else if not (isInside and ch is control)
      append ch to buffer
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