Question

我有以下变量：

$text = 'This is a sentence [url=http://site.com] that contains urls [url=javascript:alert();]';

以及用于验证网址的以下代码行：

$text = preg_replace('/\[url=([^\\]]+)]/', filter_var('$1', FILTER_VALIDATE_URL), $text);

输出：

“这是一个包含网址的句子”

如何将以下内容作为输出？

“这是包含网址”

的句子http://site.com

Answer 1

$text = 'This is a sentence [url=http://site.com] that contains urls [url=javascript:alert();]';
$text = preg_replace_callback(
    '/\[url=([^\\]]+)]/',
    function ($url) {
        $clean = filter_var($url[1], FILTER_VALIDATE_URL);
        if ($clean) {
            return '<a href="' . $clean . '">' . $clean . '</a>';
        } else {
            return "";
        }
    },
    $text
);

请参阅http://codepad.viper-7.com/Hwgq90

Answer 2

<?php

$text = 'This is a sentence [url=http://site.com] that contains urls [url=javascript:alert();]';


function url($url){
    if(filter_var($url, FILTER_VALIDATE_URL))
        return $url;
    return '';
}


$text = preg_replace('/\[url=([^\\]]+)]/e', 'url("$1")', $text);


echo $text;

更改正则表达式模式匹配函数返回值

2 个答案: