php - preg_replace，值显示为 $1.. 但使用了修改器 "e"

Question

我试图用 preg_replace 制作一个 bbCode 系统，它运行一个函数，但由于某种原因，输出仅为 1 美元，即使我添加了 'e' 修饰符。

任何想法如何解决这个问题？

源代码:

            function bbcode($value){
                $pattern[0] = "/\{register\}(.*?)\{\/register\}/sie";
                $replace[0] = register('$1'); //"<pre>$1</pre>";
                $bbcoded = preg_replace($pattern, $replace, $value);
                return $bbcoded;
            }

            function register($var) {
                echo $var;
            }

            $content = '
                {b}text{/b}
                {register}
                    "register" = "setting",
                    "name" = "title",
                    "order" = "1"
                {/register}
                some text
                {register}
                    "register" = "setting",
                    "name" = "url",
                    "order" = "3"
                {/register}
                some more {b}t{/b}e{b}xt{/b}
                {register}
                    "register" = "setting",
                    "name" = "description",
                    "order" = "2"
                {/register}
                text text
            ';

            $content = bbcode($content);

看看它在行动: http://sandbox.onlinephpfunctions.com/code/08459c4fec1919e6089d3849e154ba7a83900fd7

Answer 1

您应该使用 preg_replace_callback() 相反:

 $pattern = "/\{register\}(.*?)\{\/register\}/si"; // NO /e modifier!
 $bbcoded = preg_replace_callback($pattern, 'register', $value);

 function register($var) 
 {
     // $var[0] is the whole match
     // $var[1] is the part between {register} and {/register}

     // whatever you return here will be the replacement string
     return $var[0];
 }

preg_replace_callback()的第二个参数是 callable在找到的每个匹配项上运行，其返回值用作替换。

更新
/e 的语法替换将是这个顺便说一句(整个事情应该是一个字符串):

$bbcoded = preg_replace_callback($pattern, "register('$1')", $value);