用数组中的变量替换字符串文本 - replacing string text with variables from an array

replacing string text with variables from an array

我正在制作一款游戏的通知系统。

我决定将消息存储为带有'变量'的字符串，设置为通过数组接收的数据替换。

消息示例:

This notification will display !num1 and also !num2

我从我的查询中收到的数组看起来像:

[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
    )

我想做的是用数组(11,23)中的值替换!num1和!num2。

消息在message_tbl的查询中被INNER join。我想棘手的部分是num3，它被存储为null。

我试图将所有不同类型消息的所有通知仅存储在2个表中。

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
    )

在PHP中是否有一种方法可以成功地将num(x)替换为数组中的正确值?

您可以使用正则表达式和自定义回调来完成此操作，如下所示:

$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']);

您可以从这个演示中看到输出:

This notification will display 11 and also 23

此处:

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