php - preg_quote() 错误或奇怪的行为？

Question

我使用 preg_quote() 来转义一些特殊字符。但是在下面的测试中，有一些我无法理解的东西。

为什么“日期：1111aaa”不匹配？？？

<?php
$PregArray = ['date:1111aaa', ':222aaa', '@odia tvled'];


$array['attract (step-date:1111aaa)'] = 'OK';
$array['type (step-date:222aaa)'] = 'OK';
$array['@odia tvled'] = 'OK';


foreach ($PregArray as $key_1 => $val_1) {
    echo "\n--------------";
    foreach ($array as $key_2 => $val_2) {
        if (preg_match("~$val_1~", preg_quote($key_2))) {
            echo "\nOK => $val_1 - ".preg_quote($key_2);
            break;
            }
        else {
            echo "\nNOK !!! => $val_1 - ".preg_quote($key_2);
            }
        }
    }

http://sandbox.onlinephpfunctions.com/code/9ef36e2b010ac1f4f0eaf08881004892bd0cd9c4

Answer 1

您preg_quote在错误的地方使用它，它旨在转义模式中的字符，而不是测试字符串（请参阅手册）。将您的代码更改为此，它工作正常：

foreach ($PregArray as $key_1 => $val_1) {
    echo "\n--------------";
    foreach ($array as $key_2 => $val_2) {
        if (preg_match('~' . preg_quote($val_1) . '~', $key_2)) {
            echo "\nOK => $val_1 - $key_2";
            break;
            }
        else {
            echo "\nNOK !!! => $val_1 - $key_2";
            }
        }
    }

输出：

--------------
OK => date:1111aaa - attract (step-date:1111aaa)
--------------
NOK !!! => :222aaa - attract (step-date:1111aaa)
OK => :222aaa - type (step-date:222aaa)
--------------
NOK !!! => @odia tvled - attract (step-date:1111aaa)
NOK !!! => @odia tvled - type (step-date:222aaa)
OK => @odia tvled - @odia tvled

3v4l.org 上的演示