獲取具有多個preg_match的字符串的最後部分

Question

Spotify有兩種使用url /標識符的方法。我想要得到的字符串的最後部分下方（標識）

example url's: 
a. https://play.spotify.com/artist/6mdiAmATAx73kdxrNrnlao 
b. spotify:artist:6mdiAmATAx73kdxrNrnlao 

不能讓下面的代碼的工作，這樣我就可以把它添加更多選項後也是如此。我首先嚐試了basename，但顯然這不適用於'：'。

$str = "https://play.spotify.com/artist/6mdiAmATAx73kdxrNrnlao"; 
or: 
$str = "spotify:artist:6mdiAmATAx73kdxrNrnlao"; 

if (
    preg_match('artist/([a-zA-Z0-9]{22})/', $str, $re) || 
    preg_match('artist:([a-zA-Z0-9]{22})/', $str, $re) 

) { 
    $spotifyId = $re[1]; 
} 

任何幫助表示讚賞！

Answer 1

試試這個爲有斜線的網址。如果Spotify的字符串使用冒號（:）簡單地切換/到:在explode()功能：

// your url 
$url = "https://play.spotify.com/artist/6mdiAmATAx73kdxrNrnlao/blah/blah/blah"; 

// get path only 
$path = parse_url($url)['path']; 

// seperate by forward slash 
$parts = explode('/', $path); 

// go through them and find string with 22 characters 
$id = ''; 
foreach ($parts as $key => $value) { 
    if (strlen($value) === 22) { 
     // found it, now store it 
     $id = $value; 
     break; 
    } 
} 

一個有用功能的粗略樣品將是如下：

function getSpotifyId($spotifyUrl) { 
    // check for valid url 
    if (!filter_var($spotifyUrl, FILTER_VALIDATE_URL)) { 
     // split using colon 
     $parts = explode(':', parse_url($spotifyUrl)['path']); 
    } elseif (filter_var($spotifyUrl, FILTER_VALIDATE_URL)) { 
     // split using forward slash 
     $parts = explode('/', parse_url($spotifyUrl)['path']); 
    } 

    // loop through segments to find id of 22 chars 
    foreach ($parts as $key => $value) { 
     // assuming id will always be 22 characters 
     if (strlen($value) === 22) { 
      // found it, now return it 
      return $value; 
     } 
    } 
    return false; 
} 

$id1 = getSpotifyId('http://localhost/xampp/web_development/6mdiAmATAx73kdxrNrnlao/stack.php'); 
$id2 = getSpotifyId('spotify:artist:6mdiAmATAx73kdxrNrnlao'); 
$id3 = getSpotifyId('My name is tom'); 

結果：

$id1 = '6mdiAmATAx73kdxrNrnlao'

$id2 = '6mdiAmATAx73kdxrNrnlao'

$id3 = false