C#和JavaScript 计算时间差

     

      最近做一个人力资源系统，用户要求计算出来的合同期限是几年几天，试用期期限是几月几天。第一次遇到这样的问题，唉。做起来还是有点麻烦。要考虑平年闰年、大月小月等因素。后来找到了个好点的方法，写下来给大家分享下。c#和javascript两种版本

      首先写javascript的解决方法：

<script language=javascript>//测试代码sDate=new Date("2009/09/09");eDate=new Date("2010/03/02");alert(dateDiffToYearDay(sDate,eDate));alert(dateDiffToMonthDay(sDate,eDate));//日期相加计算，返回计算后的Date function dateAdd(strInterval, NumDay, dtDate) { var dtTmp = new Date(dtDate); if (isNaN(dtTmp)) return null; switch (strInterval) { case "s": return new Date(Date.parse(dtTmp) + (1000 * NumDay)); case "n": return new Date(Date.parse(dtTmp) + (60000 * NumDay)); case "h": return new Date(Date.parse(dtTmp) + (3600000 * NumDay)); case "d": return new Date(Date.parse(dtTmp) + (86400000 * NumDay)); case "w": return new Date(Date.parse(dtTmp) + ((86400000 * 7) * NumDay)); case "m": return new Date(dtTmp.getFullYear(), (dtTmp.getMonth()) + NumDay, dtTmp.getDate(), dtTmp.getHours(), dtTmp.getMinutes(), dtTmp.getSeconds()); case "y": return new Date((dtTmp.getFullYear() + NumDay), dtTmp.getMonth(), dtTmp.getDate(), dtTmp.getHours(), dtTmp.getMinutes(), dtTmp.getSeconds()); } }//计算两个Date相差几年几天,返回n年n天function dateDiffToYearDay(sDate,eDate){n=10000;nYear=0;nDay=0;for(i=1;i<n;i++){nextYear=dateAdd('y', i, sDate);nextYear=dateAdd('d', -1, nextYear);if(nextYear>eDate){//超过了结束日期，倒回一年计算天数nextYear=dateAdd('y', i-1, sDate);nextYear=dateAdd('d', -1, nextYear);nDay=parseInt(Math.abs(eDate - nextYear) / 1000 / 60 / 60 /24)break;}else{//增加1年nYear++;}}return (nYear+"年"+nDay+"天");}//计算两个Date相差几月几天,返回n月n天function dateDiffToMonthDay(sDate,eDate){n=10000;nMonth=0;nDay=0;for(i=1;i<n;i++){nextMonth=dateAdd('m', i, sDate);nextMonth=dateAdd('d', -1, nextMonth);if(nextMonth>eDate){//超过了结束日期，倒回一月计算天数nextMonth=dateAdd('m', i-1, sDate);nextMonth=dateAdd('d', -1, nextMonth);nDay=parseInt(Math.abs(eDate - nextMonth) / 1000 / 60 / 60 /24)break;}else{//增加1月nMonth++;}}return (nMonth+"月"+nDay+"天");}</script>

 

 

接下来是C#的解决方法：

 

//日期相加计算，返回计算后的Date private DateTime DateAdd(string strInterval, int NumDay, DateTime dtTemp) { switch (strInterval) { case "s": return dtTemp.AddSeconds(NumDay); case "n": return dtTemp.AddMinutes(NumDay); case "h": return dtTemp.AddHours(NumDay); case "d": return dtTemp.AddDays(NumDay); case "m": return dtTemp.AddMonths(NumDay); case "y": return dtTemp.AddYears(NumDay); default: return DateTime.Parse("1900-01-01"); } } //计算两个Date相差几年几天,返回n年n天 private string DateDiffToYearDay(DateTime sDate, DateTime eDate) { if (sDate > eDate) return "0年0天"; int n = 10000; int nYear = 0; int nDay = 0; for (int i = 1; i < n; i++) { DateTime nextYear = DateAdd("y", i, sDate); nextYear = DateAdd("d", -1, nextYear); if (nextYear > eDate) { //超过了结束日期，倒回一年计算天数 nextYear = DateAdd("y", i - 1, sDate); nextYear = DateAdd("d", -1, nextYear); nDay = (eDate - nextYear).Days; break; } else { //增加1年 nYear++; } } return (nYear + "/" + nDay); } //计算两个Date相差几月几天,返回n月n天 private string dateDiffToMonthDay(DateTime sDate, DateTime eDate) { if (sDate > eDate) return "0月0天"; int n = 10000; int nYear = 0; int nDay = 0; for (int i = 1; i < n; i++) { DateTime nextYear = DateAdd("m", i, sDate); nextYear = DateAdd("d", -1, nextYear); if (nextYear > eDate) { //超过了结束日期，倒回一年计算天数 nextYear = DateAdd("m", i - 1, sDate); nextYear = DateAdd("d", -1, nextYear); nDay = (eDate - nextYear).Days; break; } else { //增加1年 nYear++; } } return (nYear + "/" + nDay ); }

 

 

OK！可以调方法做下测试了。