实现身份证校验

来源：互联网发布：淘米网络私有化编辑：程序博客网时间：2024/05/16 15:40
简单的了解一下省份证的组成：第1~6位是地址码，第7~14是出生年月日，第15~17位是顺序码，第18位是校验码。
简单分析一下18位的含义：第1~2位行政代码，第3~4位低级行政区划分代码，第5~6位县级行政区分代码，第7~10位、第11~12位、第13~14位分别表示出生年、月、日，第15~17位是顺序码，第17位还可以判断一个人的性别，第18位是校验码。
具体详情请点击：1、第二代身份证号码编排规则 2、身份证号码编排规则及校验算法
具体校验实现代码如下：
import java.util.Arrays;import java.util.Calendar;import java.util.regex.Matcher;import java.util.regex.Pattern;public class IDCardRegex {    /**     * 省，直辖市代码表： { 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"}     */    protected String codeAndCity[][] = {{"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"}};    private static String cityCode[] = { "11", "12", "13", "14", "15", "21",            "22", "23", "31", "32", "33", "34", "35", "36", "37", "41", "42",            "43", "44", "45", "46", "50", "51", "52", "53", "54", "61", "62",            "63", "64", "65", "71", "81", "82", "91" };    /**     * 校验省份证正确返回码     */    public static String isCardID = "true";    private static String error_cityCode = "您输入的身份证地址编码不对，请核对后再输";    private static String error_year = "您输入的年份有误，请核对后再输";    private static String error_day = "您输入的生日日期有误，请核对后再输";    private static String error_legal = "您的还没到合法年龄再等等吧！";    private static String error_cardId = "您输入的身份证有误，请核对后再输";    private static String error_cardId_num = "您输入的身份证不足18位，请核对后再输";    private static String error_failure = "您输入的身份证号可能已失效，请核对后再输";    private static String mouth_day = "(((0[13578]|1[02])(0[1-9]|[12][0-9]|3[01]))|((0[469]|11)(0[1-9]|[12][0-9]|30))|(02(0[1-9]|[1][0-9]|2[0-9])))";    // 日期判断 平年和闰年 2月的天数    private static String isCard1 = "^[1-9]\\d{5}[1-9]\\d{3}" + mouth_day + "\\d{3}([0-9]|X|x)$";// 3 比较精确的    private String isCard2 = "^[1-9]\\d{5}[1-9]\\d{3}((0\\d)|(1[0-2]))(([0|1|2]\\d)|3[0-1])\\d{3}([0-9]|X|x)$";// 1    // 身份证验证 15 18    private static String isCarId3 = "^[1-9]\\d{7}((0\\d)|(1[0-2]))(([0|1|2]\\d)|3[0-1])\\d{3}$|^[1-9]\\d{5}[1-9]\\d{3}((0\\d)|(1[0-2]))(([0|1|2]\\d)|3[0-1])\\d{3}([0-9]|X|x)$";// 2    // 每位加权因子    private static int power[] = { 7, 9, 10, 5, 8, 4, 2, 1, 6, 3, 7, 9, 10, 5,8, 4, 2 };    // 第18位校检码    private static String verifyCode[] = { "1", "0", "X", "9", "8", "7", "6", "5", "4", "3", "2" };    public static void main(String[] args) {        String card1 = "44522119940201887X";// 虚拟的        String card2 = "15010219810707645X";// 虚拟的        String temp =card1;        System.out.println(temp);        isCard(temp);    }    /**     * 判断省份证是否符合身份证规则     * @param card String     * @return String     */    public static String isCard(String card) {        if ("".equals(card) || " ".equals(card)) {            System.out.println(error_cardId);            return error_cardId;        }else if (card.length() < 18) {            System.out.println(error_cardId_num);            return error_cardId_num;        }else if (card.length() != 18) {            System.out.println(error_cardId);            return error_cardId;        }else if (!card.matches(isCard1)) {            System.out.println(error_cardId);            return error_cardId;        }        String IdCard2 = card.substring(0, 2);// 获取前两位地址码        if (!Arrays.asList(cityCode).contains(IdCard2)) {            System.out.println(error_cityCode);            return error_cityCode;        }        String IdCard17 = card.substring(0, 17);// 获取前17位        String idCard18 = card.substring(17, 18);// 获取校验码        String IdCardyear = card.substring(6, 10);// 年        String IdCardmouth = card.substring(10, 12);// 月        String IdCardday = card.substring(12, 14);// 日        // 获取当前年        Calendar ca = Calendar.getInstance();        int cardyear = (int) Integer.parseInt(IdCardyear);        int year = ca.get(Calendar.YEAR);// 获取年份        if (year < cardyear) { // 判断年上限            System.out.println(error_year);            return error_year;        } else if (year - cardyear < 18) {// 判断合法年龄            System.out.println(error_legal);            return error_legal;        } else if (year - cardyear > 120) {// 判断人是否还在            System.out.println(error_failure);            return error_failure;        }        int cardday = (int) Integer.parseInt(IdCardday);        if ("02".equals(IdCardmouth)) {            if (cardday>28 && !isRunnian(cardyear)) {                System.out.println(error_day);                return error_day;            }        }        char[] cardChars = IdCard17.toCharArray();        int[] cardInt = new int[IdCard17.length()];        for (int i = 0; i < cardChars.length; i++) {            cardInt[i] = Integer.parseInt(String.valueOf(cardChars[i]));        }        int power = getPowerSum(cardInt);        String checkCode = verifyCode[power % 11];        System.out.println(checkCode);        if (idCard18.equals(checkCode)) {            System.out.println(isCardID);            return isCardID;        }        System.out.println(error_cardId);        return error_cardId;    }    /**     * 判断 平闰年     * @param year     * @return     */    public static boolean isRunnian(int year) {        return (year % 4 == 0 && year % 100 != 0 || year % 400 == 0)?true:false;    }    /**     * 将身份证的每位和对应位的加权因子相乘之后，再得到和值     * @param bit     * @return     */    private static int getPowerSum(int[] bit) {        int sum = 0;        if (power.length != bit.length) {            return sum;        }        for (int i = 0; i < bit.length; i++) {            for (int j = 0; j < power.length; j++) {                if (i == j) {                    sum = sum + bit[i] * power[j];                }            }        }        return sum;    }}
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