我如何过滤数组并匹配 return

Question

我想要 return 项 "like" 其他东西。我试过array_filter，但不能正确使用。

这是我试过的。期望的输出是

one.php2000565, one.php999.php . Array([0] => one.php2000565[1] => two.php[2] => three.php[3] => one.php999.php[4] => four.php)

$search_text = 'one.php';
array_filter($array, function($a) use ($search_text) {
    return ( strpos($a, $search_text) !== false );
});

Array([0] => one.php2000565[1] => two.php[2] => three.php[3] => one.php999.php[4] => four.php)

$search_text = 'one.php';
array_filter($array, function($a) use ($search_text) {
     return ( strpos($a, $search_text) !== false );
});

Answer 1

你可以试试array_filter:

$search_text = 'one.php';

array_filter($yourArray, function($el) use ($search_text) {
       return ( strpos($el, $search_text) !== false );
});

Answer 2

$res = array_filter($files, function($files) use ($search_program) {
    return ( strpos($files, $search_program) !== false );
});
print_r($res);

Answer 3

您可以尝试以下解决方案：-

$example    = array([0] => one.php2000565[1] => two.php[2] => three.php[3] => one.php999.php[4] => four.php);
$searchword = 'one.php';
$matches    = array_filter($example, function($var) use ($searchword) { return preg_match("/\b$searchword\b/i", $var); });

可能对你有帮助。

Answer 4

您没有分配 array_filter 的结果。 PHP 的 array_filter returns修改后的数组。所以只需使用：

$array = array_filter($array, 
   function($a) use ($search_text) {
       return ( 
           strpos($a, $search_text) !== false 
       );
   }
)

Answer 5

正如我评论的那样，OP 代码在 array_filter 下工作正常，只需要在变量中分配过滤后的值。但是我的解决方案是 array_filter 的替代方案，您可以像

一样使用 preg_grep

$res = preg_grep("/$search_text/",$array);
print_r($res);

我如何过滤数组并匹配 return

how can i filter an array and return matches

php

arrays

strpos

array-filter