身份证号的函数验证

身份证号的函数验证

/^(\d{15}$|^\d{18}$|^\d{17}(\d|X|x))$/  一个比较简单的非严格的正则验证身份证号

1
//定义一个全国地区的对象 2 var aCity={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"}; 3 function isCardID(sId){ 4 var iSum=0 ; 5 var info="" ; 6 if(!/^\d{17}(\d|x)$/i.test(sId)) return layer.msg("你输入的身份证长度或格式错误",{icon:2}); 7 sId=sId.replace(/x$/i,"a"); 8 if(aCity[parseInt(sId.substr(0,2))]==null) return layer.msg("你的身份证地区非法",{icon:2}); 9 sBirthday=sId.substr(6,4)+"-"+Number(sId.substr(10,2))+"-"+Number(sId.substr(12,2)); 10 var d=new Date(sBirthday.replace(/-/g,"/")) ; 11 if(sBirthday!=(d.getFullYear()+"-"+ (d.getMonth()+1) + "-" + d.getDate()))return layer.msg("身份证上的出生日期非法",{icon:2}); 12 for(var i = 17;i>=0;i --) iSum += (Math.pow(2,i) % 11) * parseInt(sId.charAt(17 - i),11) ; 13 if(iSum%11!=1) return layer.msg("你输入的身份证号非法",{iocn:2}) 14 //aCity[parseInt(sId.substr(0,2))]+","+sBirthday+","+(sId.substr(16,1)%2?"男":"女");//此次还可以判断出输入的身份证号的人性别 15 return true; 16 }

 

posted @ 2017-05-19 15:47  努力努力要努力  阅读(613)  评论(0编辑  收藏  举报