Home  >  Article  >  Web Front-end  >  jQuery code to get age based on date of birth

jQuery code to get age based on date of birth

小云云
小云云Original
2018-02-27 11:20:561977browse


本文主要和大家分享jquery 根据出生年月日得到年龄,希望能帮助到大家。
jquery 根据出生年月日得到年龄
function byage(strBirthday ){
		   var returnAge;  
		    var strBirthdayArr=strBirthday.split("-");  
		    var birthYear = strBirthdayArr[0];  
		    var birthMonth = strBirthdayArr[1];  
		    var birthDay = strBirthdayArr[2];  
		      
		    d = new Date();  
		    var nowYear = d.getFullYear();  
		    var nowMonth = d.getMonth() + 1;  
		    var nowDay = d.getDate();  
		      
		    if(nowYear == birthYear){  
		        returnAge = 0;//同年 则为0岁  
		    }  
		    else{  
		        var ageDiff = nowYear - birthYear ; //年之差  
		        if(ageDiff > 0){  
		            if(nowMonth == birthMonth) {  
		                var dayDiff = nowDay - birthDay;//日之差  
		                if(dayDiff < 0)  
		                {  
		                    returnAge = ageDiff - 1;  
		                }  
		                else  
		                {  
		                    returnAge = ageDiff ;  
		                }  
		            }  
		            else  
		            {  
		                var monthDiff = nowMonth - birthMonth;//月之差  
		                if(monthDiff < 0)  
		                {  
		                    returnAge = ageDiff - 1;  
		                }  
		                else  
		                {  
		                    returnAge = ageDiff ;  
		                }  
		            }  
		        }  
		        else  
		        {  
		            returnAge = -1;//返回-1 表示出生日期输入错误 晚于今天  
		        }  
		    }  
		      
		    return returnAge;//返回周岁年龄
	}

Related recommendations:

JS method to get age based on date of birth

The above is the detailed content of jQuery code to get age based on date of birth. For more information, please follow other related articles on the PHP Chinese website!

Statement:
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