Question

我有一个我正在解析的文件而我正在尝试用$mail["email_from"] = "test@example.com";替换$mail["email_from"] = request("email");，（这意味着我想要替换$mail["email_from"]处的所有行。最后开始;）这里是我的preg_replace：

$email = "$mail[\"email_from\"] = request(\"email\")";
$newcontent = preg_replace("/\$mail[\"email_from\"](.+);/",$email,$content);

我的代码中出现了什么错误？以及如何解决它？非常感谢

DEMO

Answer 1

$email = "$mail[\"email_from\"] = request(\"email\")";
         ^---double-quoted string
          ^^^^^---array reference

您可能需要

$email = "\$mail[\"email_from\"] = request(\"email\")";
          ^--escape the $

Answer 2

使用好的引号并转义所需的所有字符后，这可以工作：

$email = '$mail["email_from"] = "test@example.com";';
$replacement = '$mail["email_from"] = request("email");';
$newContent = preg_replace('/\\$mail\\[\\"email_from\\"\\](.+);/i', $replacement, $email); 
echo $newContent; //$mail["email_from"] = request("email");

Answer 3

使用^和$指定行的开头和结尾。需要转发$，[和]等特殊字符。

<?php
$content = '$mail["email_from"] = "test@example.com";';
$email = '$mail["email_from"] = request("email");';
$newcontent = preg_replace('/^\$mail\["email_from"\] =.+;$/',$email,$content); 

echo $newcontent . "\n";

输出：

$mail["email_from"] = request("email");

preg_replace里面的PHP文件

3 个答案: