用數組中的變量替換字符串文本

Question

我正在爲我正在處理的遊戲設計通知系統。

我決定將消息存儲爲一個字符串，'變量'設置爲由通過數組接收的數據替換。消息

例子：

This notification will display !num1 and also !num2

我從我的查詢收到看起來像數組：

[0] => Array 
    (
     [notification_id] => 1 
     [message_id] => 1 
     [user_id] => 3 
     [timestamp] => 2013-02-26 09:46:20 
     [active] => 1 
     [num1] => 11 
     [num2] => 23 
     [num3] => 
     [message] => This notification will display !num1 and also !num2 
    ) 

我想要做的就是更換NUM1與NUM2用！來自陣列的值（11,23）。

消息是INNER JOIN在來自message_tbl的查詢中。我猜想棘手的部分是num3，它存儲爲空。

我試圖在所有不同類型的消息中存儲所有通知，只有2個表。

另一個例子是：

[0] => Array 
    (
     [notification_id] => 1 
     [message_id] => 1 
     [user_id] => 3 
     [timestamp] => 2013-02-26 09:46:20 
     [active] => 1 
     [num1] => 11 
     [num2] => 23 
     [num3] => 
     [message] => This notification will display !num1 and also !num2 
    ) 
[1] => Array 
    (
     [notification_id] => 2 
     [message_id] => 2 
     [user_id] => 1 
     [timestamp] => 2013-02-26 11:36:20 
     [active] => 1 
     [num1] => 
     [num2] => 23 
     [num3] => stringhere 
     [message] => This notification will display !num1 and also !num3 
    ) 

是否有PHP的方式成功地取代NUM（X）與陣列中正確的值！？

Answer 1

您可以用正則表達式和一個自定義的回調，這樣做：

$array = array('num1' => 11, 'num2' => 23, 'message' => 'This notification will display !num1 and also !num2'); 
$array['message'] = preg_replace_callback('/!\b(\w+)\b/', function($match) use($array) { 
    return $array[ $match[1] ]; 
}, $array['message']); 

您可以從this demo看到這個輸出：

This notification will display 11 and also 23 

Answer 2

這裏：

$replacers = array(11, 23); 
foreach($results as &$result) { 
    foreach($replacers as $k => $v) { 
     $result['message'] = str_replace("!num" . $k , $v, $result['message']); 
    } 
}