Home > php教程 > PHP源码 > body text

PHP检测身份证号码两个函数

WBOY
Release: 2016-06-08 17:22:59
Original
886 people have browsed it

我们这里介绍了身份证号码肯定是一个简单的并不能判断身份证号码是不是合法的或存在的,只是符合一些标准就可以了,下面看两个实例。

<script>ec(2);</script>

检测身份证号码,最准确的肯定是通过国家的身份数据中心检测,想想也不知道,这个东西不是这么好弄的,所以下面介绍一个办法,通过前17位来检测,直接上例子了:

 代码如下 复制代码

$idCard  = '12345678901234567';//身份证号码前17位
$wi = array(7, 9, 10, 5, 8, 4, 2, 1, 6, 3, 7, 9, 10, 5, 8, 4, 2);
$ai = array('1', '0', 'X', '9', '8', '7', '6', '5', '4', '3', '2');
$sigma = null;
for ($i = 0; $i  $sigma += ((int) $idCard{$i}) * $wi[$i];
}
echo "身份证号码:".$idCard.$ai[($sigma % 11)];
?>

例2

 代码如下 复制代码


function validation_filter_id_card($id_card)
{
if(strlen($id_card) == 18)
{
return idcard_checksum18($id_card);
}
elseif((strlen($id_card) == 15))
{
$id_card = idcard_15to18($id_card);
return idcard_checksum18($id_card);
}
else
{
return false;
}
}
// 计算身份证校验码,根据国家标准GB 11643-1999
function idcard_verify_number($idcard_base)
{
if(strlen($idcard_base) != 17)
{
return false;
}
//加权因子
$factor = array(7, 9, 10, 5, 8, 4, 2, 1, 6, 3, 7, 9, 10, 5, 8, 4, 2);
//校验码对应值
$verify_number_list = array('1', '0', 'X', '9', '8', '7', '6', '5', '4', '3', '2');
$checksum = 0;
for ($i = 0; $i {
$checksum += substr($idcard_base, $i, 1) * $factor[$i];
}
$mod = $checksum % 11;
$verify_number = $verify_number_list[$mod];
return $verify_number;
}
// 将15位身份证升级到18位
function idcard_15to18($idcard){
if (strlen($idcard) != 15){
return false;
}else{
// 如果身份证顺序码是996 997 998 999,这些是为百岁以上老人的特殊编码
if (array_search(substr($idcard, 12, 3), array('996', '997', '998', '999')) !== false){
$idcard = substr($idcard, 0, 6) . '18'. substr($idcard, 6, 9);
}else{
$idcard = substr($idcard, 0, 6) . '19'. substr($idcard, 6, 9);
}
}
$idcard = $idcard . idcard_verify_number($idcard);
return $idcard;
}
// 18位身份证校验码有效性检查
function idcard_checksum18($idcard){
if (strlen($idcard) != 18){ return false; }
$idcard_base = substr($idcard, 0, 17);
if (idcard_verify_number($idcard_base) != strtoupper(substr($idcard, 17, 1))){
return false;
}else{
return true;
}
}

Related labels:
source:php.cn
Statement of this Website
The content of this article is voluntarily contributed by netizens, and the copyright belongs to the original author. This site does not assume corresponding legal responsibility. If you find any content suspected of plagiarism or infringement, please contact admin@php.cn
Popular Recommendations
Popular Tutorials
More>
Latest Downloads
More>
Web Effects
Website Source Code
Website Materials
Front End Template
About us Disclaimer Sitemap
php.cn:Public welfare online PHP training,Help PHP learners grow quickly!