用数组中的变量替换字符串文本

Question

我正在为我正在处理的游戏设计通知系统。

我决定将消息存储为一个字符串，'变量'设置为由通过数组接收的数据替换。消息

例子：

This notification will display !num1 and also !num2

我从我的查询收到看起来像数组：

[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 
    ) 

我想要做的就是更换NUM1与NUM2用！来自阵列的值（11,23）。

消息是INNER JOIN在来自message_tbl的查询中。我猜想棘手的部分是num3，它存储为空。

我试图在所有不同类型的消息中存储所有通知，只有2个表。

另一个例子是：

[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）与阵列中正确的值！？

Answer 1

您可以用正则表达式和一个自定义的回调，这样做：

$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']); 

您可以从this demo看到这个输出：

This notification will display 11 and also 23 

Answer 2

这里：

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