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考虑以下包含混淆的电子邮件地址的脚本，以及一个尝试*****使用正则表达式模式匹配替换它们的函数。我的脚本试图捕捉这些词："at", "a t", "a.t", "@"后跟一些文本（任何域名），然后是"dot" "." "d.o.t"，然后是 TLD。

输入：

$str[] = 'dsfatasdfasdf asd dsfasdf dsfdsf@hotmail.com'; 
$str[] = 'I live at school where My address is dsfdsf@hotmail.com'; 
$str[] = 'I live at school. My address is dsfdsf@hotmail.com'; 
$str[] = 'at school my address is dsfdsf@hotmail.com'; 
$str[] = 'dsf a t asdfasdf asd dsfasdf dsfdsf@hotmail.com'; 
$str[] = 'd s f d s f a t h o t m a i l . c o m';

function clean_text($text){
    $pattern = '/(\ba[ \.\-_]*t\b|@)[ \.\-_]*(.+)[ \.\-_]*(d[ \.\-_]*o[ \.\-_]*t|\.)[ \.\-_]*(c[ \.\-_]*o[ \.\-_]*m|n[ \.\-_]*e[ \.\-_]*t|o[ \.\-_]*r[ \.\-_]*g|([a-z][ \.\-_]*){2,3}[a-z]?)/iU'; 
    return preg_replace($pattern, '***', $text); 
}

foreach($str as $email){ 
     echo clean_text($email); 
}

预期输出： 

dsfatasdfasdf asd dsfasdf dsfdsf*** 
I live at school where My address is dsfdsf@***
I live at school. My address is dsfdsf@***
*** 
dsf *** 
d s f d s f ***

结果：

dsfatasdfasdf asd dsfasdf dsfdsf*** 
I live *** 
I live *** 
at school my address is dsfdsf****
dsf *** 
d s f d s f ***

问题： 它捕获了“at”的第一次出现，而不是最后一次出现，所以会发生以下情况：

input: 'at school my address is dsfdsf@hotmail.com'
produces: '****'
should produce: 'at school my address is dsfdsf****'

我怎样才能解决这个问题？

4

3 回答 3

2

基于 M42 的正则表达式：

代码：

$emails = array(
                'dsfatasdfasdf asd dsfasdf dsfdsf@hotmail.com'
                ,'I live at school where My address is dsfdsf@hotmail.com'
                ,'I live at school. My address is dsfdsf@hotmail.com'
                ,'at school my address is dsfdsf@hotmail.com'
                ,'dsf a t asdfasdf asd dsfasdf dsfdsf@hotmail.com'
                ,'d s f d s f a t h o t m a i l . c o m'
                );

foreach($emails as $email)
{
    $found = preg_match('/(.*?)((\@|a[_. -]*t)[\w .-]*?$)/', $email, $matches);
    if($found)
    {
        echo 'Username: ' . $matches[1] . ', Domain: ' . $matches[2] . "\n";
    }
}

输出：

Username: dsfatasdfasdf asd dsfasdf dsfdsf, Domain: @hotmail.com
Username: I live at school where My address is dsfdsf, Domain: @hotmail.com
Username: I live at school. My address is dsfdsf, Domain: @hotmail.com
Username: at school my address is dsfdsf, Domain: @hotmail.com
Username: dsf a t asdfasdf asd dsfasdf dsfdsf, Domain: @hotmail.com
Username: d s f d s f , Domain: a t h o t m a i l . c o m
于 2010-07-27T19:22:19.733 回答
1

这是一个 Perl 脚本，可以适应 php 吗？

my @l = (
'dsfatasdfasdf asd dsfasdf dsfdsf@hotmail.com',
'I live at school where My address is dsfdsf@hotmail.com',
'I live at school. My address is dsfdsf@hotmail.com',
'at school my address is dsfdsf@hotmail.com',
'dsf a t asdfasdf asd dsfasdf dsfdsf@hotmail.com',
'd s f d s f a t h o t m a i l . c o m'
);

foreach(@l) {
   s/(\@|a[_. -]*t)[\w .-]*?$/****/;
   print $_,"\n";
}

输出：

dsfatasdfasdf asd dsfasdf dsfdsf****
I live at school where My address is dsfdsf****
I live at school. My address is dsfdsf****
at school my address is dsfdsf****
dsf a t asdfasdf asd dsfasdf dsfdsf****
d s f d s f ****
于 2010-07-27T18:52:02.907 回答
0 
function clean_text($text){
    $pattern = '/\w+[\w-\.]*(\@\w+((-\w+)|(\w*))\.[a-z]{2,3})/i';
    preg_match($pattern, $text, $matches);

    return (isset($matches[1])) ? str_replace($matches[1], "****", $text) : $text;
}

唯一不匹配的是你的最后一个，但你明白了。

于 2010-07-27T16:28:26.660 回答