Question

我想将电子邮件＃$％＃$ @ b@#$#$2344324.com清理为a@b.com。
我尝试过但失败了 echo filter_var（“a＃$％＃$ @ b@#$#$2344324.com”，FILTER_SANITIZE_EMAIL）; //结果：＃$％＃$ @ b @＃$＃$ 2344324.com

我需要修剪电子邮件中的特殊字符（清理以删除特殊字符）。我使用下面的代码，但我没有成功。

$string = preg_replace("/^[a-zA-Z0-9._%+-]+@(?:[a-zA-Z0-9-]+\.)+[a-zA-Z]{2,4}$/", "", "a#$%#$@b@#$#$2344324.com");

echo $string;//result: a@b.com -- unwanted characters trimmed here.

Answer 1

此处已有基于RFC的解决方案：http://fightingforalostcause.net/misc/2006/compare-email-regex.php

function is_valid_email_address($email){

        $qtext = '[^\\x0d\\x22\\x5c\\x80-\\xff]';

        $dtext = '[^\\x0d\\x5b-\\x5d\\x80-\\xff]';

        $atom = '[^\\x00-\\x20\\x22\\x28\\x29\\x2c\\x2e\\x3a-\\x3c'.
            '\\x3e\\x40\\x5b-\\x5d\\x7f-\\xff]+';

        $quoted_pair = '\\x5c[\\x00-\\x7f]';

        $domain_literal = "\\x5b($dtext|$quoted_pair)*\\x5d";

        $quoted_string = "\\x22($qtext|$quoted_pair)*\\x22";

        $domain_ref = $atom;

        $sub_domain = "($domain_ref|$domain_literal)";

        $word = "($atom|$quoted_string)";

        $domain = "$sub_domain(\\x2e$sub_domain)*";

        $local_part = "$word(\\x2e$word)*";

        $addr_spec = "$local_part\\x40$domain";

        return preg_match("!^$addr_spec$!", $email) ? true : false;
    }

使用preg_replace删除电子邮件中不需要的字符

1 个答案: