二代居民身份证更严格校验方法。

正常校验二代身份证是否合法，都只是校验位数与正则判断，但是身份证也有一个算法 根据最后一位校验位能够计算出这个身份证号码是否合法，更加严格 如下：

export function IdentityCodeValid(code) {

let city={11:"北京",12:"天津",13:"河北",14:"山西",15:"内蒙古",21:"辽宁",22:"吉林",23:"黑龙江 ",31:"上海",32:"江苏",33:"浙江",34:"安徽",35:"福建",36:"江西",37:"山东",41:"河南",42:"湖北 ",43:"湖南",44:"广东",45:"广西",46:"海南",50:"重庆",51:"四川",52:"贵州",53:"云南",54:"西藏 ",61:"陕西",62:"甘肃",63:"青海",64:"宁夏",65:"新疆",71:"台湾",81:"香港",82:"澳门",91:"国外 "};

let pass= true;

let reg = /^(^[1-9]\d{7}((0\d)|(1[0-2]))(([0|1|2]\d)|3[0-1])\d{3}$)|(^[1-9]\d{5}[1-9]\d{3}((0\d)|(1[0-2]))(([0|1|2]\d)|3[0-1])((\d{4})|\d{3}[Xx])$)$/;
if(!code || reg.test(code) === false) { //验证身份证是否有效
    pass = false;
}

else if(!/^\d{6}(18|19|20)?\d{2}(0[1-9]|1[0-2])(0[1-9]|[12]\d|3[01])\d{3}(\d|X)$/i.test(code)){
    pass = false;
}

else if(!city[code.substr(0,2)]){
    pass = false;
}
else{
    //18位身份证需要验证最后一位校验位
    if(code.length == 18){
        code = code.split('');
        //∑(ai×Wi)(mod 11)
        //加权因子
        let factor = [ 7, 9, 10, 5, 8, 4, 2, 1, 6, 3, 7, 9, 10, 5, 8, 4, 2 ];
        //校验位
        let parity = [ 1, 0, 'X', 9, 8, 7, 6, 5, 4, 3, 2 ];
        let sum = 0;
        let ai = 0;
        let wi = 0;
        for (let i = 0; i < 17; i++)
        {
            ai = code[i];
            wi = factor[i];
            sum += ai * wi;
        }
        let last = parity[sum % 11];
        if(parity[sum % 11] != code[17]){
            pass =false;
        }
    }
}
return pass;

}