搜索单数和复数形式的针

Question

I am looking for a good solution to check, whether a word (even in the singular-version) is existing in a specific string.

Example:

$search = 'users';

$array = [
    'index-users',
    'view-users',
    'create-users',
    'index-user-roles',
    'view-user-roles',
    'create-user-roles',
    'index-todos',
    'view-todos',
    'create-todos',
    'index-attributes',
    'view-attributes',
    'create-attributes',
];

I expect the result for the check for users to be true here.

Even in this example (Singular) the result should be true :

<?php

$search = 'users';

$array = [
    'view-user-roles',
    'create-user-roles',
    'index-todos',
    'view-todos',
    'create-todos',
    'index-attributes',
    'view-attributes',
    'create-attributes',
];

I thought about using a regex, but my skills are not good enough to cover the plural and singular-variants.

I tried

\b(\w*users\w*)\b

but this does not cover the singular variant. I added a short example: https://www.regextester.com/?fam=110732

Answer 1

You can achieve something like this with preg_grep()

$array = [
        'index-users',
        'view-users',
        'create-users',
        'index-user-roles',
        'view-user-roles',
        'create-user-roles',
        'index-todos',
        'view-todos',
        'create-todos',
        'index-attributes',
        'view-attributes',
        'create-attributes',
];

$search = 'use'; 

$result = preg_grep('~' . $search . '~', $array);
if (!empty($result)){
   var_dump($result);
} else {
  echo 'No match found';
}