如果它與 PHP 的特定字符串匹配，如何從數組中完全檢索 URL

Question

如何通過包含多個 URL 的數組搜索和檢索包含特定字符串（包含在變量中）的 URL？

我嘗試了幾個函數，包括 array_intersect，但匹配必須與這個函數完美匹配。

我找到的唯一方法是 preg_match_all 但我是 PHP 的新手，我不知道應該向 preg_match_all 函數添加什么確切的正則表達式和參數，你能幫我嗎？

array() {
    [0]=>
    string(45) "http://localhost/website02/wp-content/uploads/2022/12/2023-lamborghini-huracan-sterrato-1-2.jpg"
    [1]=>
    string(100) "http://localhost/website02/wp-content/uploads/2022/12/alpine-a4810-hydrogen-concept-front-corner.jpg" 
   [2]=>
   string(74) "http://localhost/website02/wp-content/uploads/2022/12/lexus-example-1.jpg"
}


$mystring = "example";

$pattern = "/$mystring/i";

if(preg_match($pattern, $arr, $matches)) {
  print_r($matches[0][1]);
  /************ I would like the result: http://localhost/website02/wp-content/uploads/2022/12/lexus-example-1.jpg *****************/
}

Answer 1

您可以試試這段代碼，這將返回匹配“example”的 url

$array = ['http://localhost/website02/wp-content/uploads/2022/12/2023-lamborghini-huracan-sterrato-1-2.jpg','http://localhost/website02/wp-content/uploads/2022/12/alpine-a4810-hydrogen-concept-front-corner.jpg','http://localhost/website02/wp-content/uploads/2022/12/lexus-example-1.jpg'];


foreach($array as $url)
{
    $baseName = basename($url);
    $str_arr = preg_split ("/\-/", $baseName);
    
        foreach($str_arr as $value)
        {
            if(preg_match('/example/i', $value))
            {
                echo $url;
            }
        }

}