计算两个日期相差的天数 js php日期 减一年

计算两个日期相差的天数

    //sDate1和sDate2是yyyy-MM-dd格式
    function dateDiff(sDate1, sDate2) {
        var aDate, oDate1, oDate2, iDays,iMonth;
        aDate = sDate1.split("-");
        oDate1 = new Date(aDate[1] + '-' + aDate[2] + '-' + aDate[0]);  //转换为yyyy-MM-dd格式
        aDate = sDate2.split("-");
        oDate2 = new Date(aDate[1] + '-' + aDate[2] + '-' + aDate[0]);

        iDays = parseInt(Math.abs(oDate1 - oDate2) / 1000 / 60 / 60 / 24); //把相差的毫秒数转换为天数
        iMonth = parseInt(Math.abs(oDate1 - oDate2) / 1000 / 60 / 60 / 24 /30); //把相差的毫秒数转换为月

        return {d:iDays,m:iMonth};  //返回相差天数
    }

    dateDiff('2017-02-02','2017-02-15');

　　

//两个日期之间所有的日期

var getDates = function(startDate, endDate) {
  var dates = [],
      currentDate = startDate,
      addDays = function(days) {
        var date = new Date(this.valueOf());
        date.setDate(date.getDate() + days);
        return date;
      };
  while (currentDate <= endDate) {
    dates.push(currentDate);
    currentDate = addDays.call(currentDate, 1);
  }
  return dates;
};


var dates = getDates(new Date(2017,12,22), new Date(2018,01,10));                                                                                                           
dates.forEach(function(date) {
  console.log(date);
});

　　

日期减一年

var a =new Date('2017-10-12');

a.setFullYear(a.getFullYear() +1);

referr: https://stackoverflow.com/questions/33070428/add-year-to-todays-date

 

当然在php里面减一年就简单了

$date = '2004-02-29';
echo date('Y-m-d',strtotime('-1 year',strtotime($date)));