How to cut all blocks that do not contain an attribute?

Question

PHP

The piece being tested

$test = '<input type="hidden" name="submitted" value="TRUE" />
      <!-- LOCALE uk -->
      <h2 class="">Required Information UK</h2>
      <!-- /LOCALE -->
<!-- LOCALE uk --><h2 class="">Required Information UK</h2><!-- /LOCALE -->
<!-- LOCALE ru -->
<h2 class="">Required Information RU</h2>
<!-- /LOCALE -->

<!-- LOCALE -->
<h2 class="">Required Information EMPTY</h2>
<!-- /LOCALE -->
<div class="content_indent" style="width: 764px; margin-bottom: 35px;">';


//Попытки :(
echo preg_replace('#<!--\s+LOCALE\s+(ru)\s+-->.*?<!--\s+/LOCALE\s+-->#sm','',$test);

It is necessary to cut out all blocks in which LOCALE does not have the ru attribute. While it turned out to cut out only everything with this attribute. How to specify inversion? (?!ru) - it won't start like that

Answer 1

Hello.
Here's a regex try slip. UPDATE
Example : Example 2

Answer 2

Isn't it easier to remove all blocks with ru preg_replace like this https://regex101.com/r/dX2wH3/1 and that 's it