嵌套（多個）字符串中PHP插值

Question

如果非要包含嵌套變量的變量，例如：

$message = "Hello $user_name, an email was send to $user_email ..."; 
$user_name = 'User Name'; 
$user_email = 'user@email.com'; 

是能夠產生的輸出，諸如：

Hello User Name, an email was send to user@email.com ... 

而不調用eval() ？

Answer 1

可以「定義」佔位符和在你需要它更換。

$message = "Hello #user_name#, an email was send to #user_email# ..."; 
$user_name = 'User Name'; 
$user_email = 'user@email.com'; 

$newMessage = str_replace(array("#user_name#", "#user_email#"), array($user_name, $user_email), $message); 

查看str_replace僅供參考。

Answer 2

是的，只需將變量$user_email和$user_name放置在$message以上即可，因此它們將首先被實例化。

$user_email = 'user@email.com'; 
$user_name = 'User Name'; 
$message = "Hello $user_name, an email was send to $user_email ..."; 

echo $message; //Will output: Hello User Name, an email was send to user@email.com ... 

編輯：讀你的反應後，您可以使用例如關閉：

$message = function($name = null, $email = null){ 
    return "Hello $name, an email was send to $email ..."; 
}; 

$user_name = 'User Name'; 
$user_email = 'user@email.com'; 

$newMessage = $message($user_name, $user_email); 

Answer 3

使用sprintf或printf功能：

$message = "Hello %s, an email was send to %s ..."; 
$user_name = 'User Name'; 
$user_email = 'user@email.com'; 

echo sprintf($message, $user_name, $user_email); 

或

printf($message, $user_name, $user_email); 

Answer 4

$user_email = 'user@email.com'; 
$user_name = 'User Name'; 
$message = "Hello " . $user_name . ", an email was send to " . $user_email . " ...";