12

我正在使用以下脚本来获取前一周的星期一（第一）和星期日（最后）：

var curr = new Date; // get current date
var first = curr.getDate() - curr.getDay() - 6; // Gets day of the month (e.g. 21) - the day of the week (e.g. wednesday = 3) = Sunday (18th) - 6
var last = first + 6; // last day is the first day + 6
var startDate = new Date(curr.setDate(first));
var endDate = new Date(curr.setDate(last));

如果上周一和周日也在同一个月，这很好用，但我今天才注意到，如果今天是 12 月而上周一是 11 月，它就行不通了。

我是个 JS 新手，还有其他方法可以获取这些日期吗？

4

8 回答 8

27

您可以通过获取本周的星期一并减去 7 天来获取上一个星期一。星期天将是前一天，所以：

var d = new Date();

// set to Monday of this week
d.setDate(d.getDate() - (d.getDay() + 6) % 7);

// set to previous Monday
d.setDate(d.getDate() - 7);

// create new date of day before
var sunday = new Date(d.getFullYear(), d.getMonth(), d.getDate() - 1);

对于 2012-12-03，我得到：

Mon 26 Nov 2012
Sun 25 Nov 2012

那是你要的吗？

// Or new date for the following Sunday
var sunday = new Date(d.getFullYear(), d.getMonth(), d.getDate() + 6);

这使

Sun 02 Dec 2012

通常，您可以通过加减年、月和日来操作日期对象。该对象将自动处理负值，例如

var d = new Date(2012,11,0)

将为 2012-11-30 创建一个日期（注意月份是从零开始的，因此 11 是 12 月）。还：

d.setMonth(d.getMonth() - 1); // 2012-10-30

d.setDate(d.getDate() - 30);  // 2012-09-30
于 2012-12-03T11:13:23.290 回答
7

如果您不想使用外部库来执行此操作，则应该使用时间戳。我创建了一个解决方案，您将从当前日期减去 60*60*24*7*1000（即 604800000，即 1 周，以毫秒为单位）并从那里开始：

var beforeOneWeek = new Date(new Date().getTime() - 60 * 60 * 24 * 7 * 1000)
  , day = beforeOneWeek.getDay()
  , diffToMonday = beforeOneWeek.getDate() - day + (day === 0 ? -6 : 1)
  , lastMonday = new Date(beforeOneWeek.setDate(diffToMonday))
  , lastSunday = new Date(beforeOneWeek.setDate(diffToMonday + 6));
于 2012-12-03T11:19:06.300 回答
4

你可以使用像 moment.js 这样的库。见subtract方法http://momentjs.com/docs/#/manipulating/subtract/

于 2012-12-03T11:01:09.467 回答
3

一些答案提到了时刻，但没有人写过这个简单的方法：

moment().day(-13) // Monday last week
moment().day(-7) // Sunday last week

.day设置一个工作日，所以今天是什么日子并不重要，只有星期很重要。

于 2016-06-12T10:45:26.693 回答
2

这是查找任何一周中的任何一天的一般解决方案。

function getParticularDayTimestamp(lastWeekDay) {
  var currentWeekMonday = new Date().getDate() - new Date().getDay() + 1;
  return new Date().setDate(currentWeekMonday - lastWeekDay);
}

console.log(getParticularDayTimestamp(7)) // for last week monday
console.log(getParticularDayTimestamp(1)) // for last week sunday
console.log(getParticularDayTimestamp(14)) // for last to last week monday
console.log(getParticularDayTimestamp(8)) // for last to last week sunday
于 2020-01-29T06:03:21.623 回答
1

使用 Moment，您可以执行以下操作

  var lastWeek          = moment().isoWeek(moment().subtract(1,'w').week());
  var mondayDifference  = lastWeek.dayOfYear() - lastWeek.weekday() + 1;
  var sundayDifference  = mondayDifference - 1;

  var lastMonday        = moment().dayOfYear(mondayDifference);
  var lastSunday        = moment().dayOfYear(sundayDifference  );
于 2016-01-21T22:53:05.657 回答
1

就这么简单。

var today = new Date();
var sunday = new Date(this.today.getFullYear(), this.today.getMonth(), this.today.getDate() - this.today.getDay());
于 2019-03-21T16:39:19.140 回答
0

在这里，您有一个多功能功能： 

 function getThe(numOfWeeks, weekday, tense, fromDate) { 
 // for instance: var lastMonday = getThe(1,"Monday","before",new Date()) 
      var targetWeekday = -1;  
      var dateAdjustment = clone(fromDate);  
      var result = clone(fromDate);

      switch (weekday) {
        case "Monday": targetWeekday = 8; break;
        case "Tuesday": targetWeekday = 2; break;
        case "Wednesday": targetWeekday = 3; break;
        case "Thursday": targetWeekday = 4; break;
        case "Friday": targetWeekday = 5; break;
        case "Saturday": targetWeekday = 6; break;
        case "Sunday": targetWeekday = 7;
      }

      var adjustment = 7 * (numOfWeeks - 1);
      if (tense == "after") adjustment = -7 * numOfWeeks;

      dateAdjustment.setDate(fromDate.getDate() - targetWeekday);
      var weekday = dateAdjustment.getDay();

      result.setDate(fromDate.getDate() - weekday - adjustment);
      result.setHours(0,0,0,0);
      return result;
  }

您可以在下一篇文章中找到“clone(obj)”函数：https ://stackoverflow.com/a/728694/6751764

于 2016-09-05T09:10:33.817 回答