Preg 替换为 mysql
好的，这就是我所拥有的，但每当我这样做时，我都会返回数据库的第一条记录：
<?php
function getInfo($id,$slot){
if(!$id){ return '<b>Error</b> Id Not Returned. Please contact [email protected] for more information.'; }
$mm = mysql_query("SELECT * FROM `users` WHERE `id`='".$id."'");
$mma = mysql_fetch_assoc($mm);
$p = $mma[$slot];
return $p;
//return $id; <- Debug (Returns ID given)
}
$post = preg_replace("/\[CallName]([^]]+)\[\/CallName\]/", getInfo('\\1',"fullname"), $post);
?>
Okay so heres what I have but whenever I do this, I am returned with the first record of the database:
<?php
function getInfo($id,$slot){
if(!$id){ return '<b>Error</b> Id Not Returned. Please contact [email protected] for more information.'; }
$mm = mysql_query("SELECT * FROM `users` WHERE `id`='".$id."'");
$mma = mysql_fetch_assoc($mm);
$p = $mma[$slot];
return $p;
//return $id; <- Debug (Returns ID given)
}
$post = preg_replace("/\[CallName]([^]]+)\[\/CallName\]/", getInfo('\\1',"fullname"), $post);
?>
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我很惊讶你得到了任何返回的东西......你的 preg_replace 模式应该给你一个错误（你没有转义一些括号）。但无论如何，您不能像这样将捕获的组传递给您的函数。您必须使用“e”模式修饰符（请参阅 preg_replace 手册中的示例#4
I'm surprised you're getting anything returned...your preg_replace pattern should be giving you an error (you aren't escaping some of your brackets). But anyways, you can't pass a captured group to your function like that. You have to use the "e" pattern modifier (look at example #4 in the preg_replace manual