如何在JavaScript中获取前8个星期天？

我正在尝试用JavaScript获取之前的8个星期天。这不仅不起作用，而且看起来很笨重。我觉得用阵列的东西会更干净。我做错了什么

var now = new Date();
var today = new Date(now.getFullYear(), now.getMonth(), now.getDate());
var lastSunday = new Date(today.setDate(today.getDate() - today.getDay()));
lastSunday = lastSunday.format("mm/dd/yyyy");

var lastSunday2 = new Date(today.setDate(today.getDate() - today.getDay() - 7));
lastSunday2 = lastSunday2.format("mm/dd/yyyy");
var lastSunday3 = new Date(today.setDate(today.getDate() - today.getDay() - 14));
lastSunday3 = lastSunday3.format("mm/dd/yyyy");
var lastSunday4 = new Date(today.setDate(today.getDate() - today.getDay() - 21));
lastSunday4 = lastSunday4.format("mm/dd/yyyy");
var lastSunday5 = new Date(today.setDate(today.getDate() - today.getDay() - 28));
lastSunday5 = lastSunday5.format("mm/dd/yyyy");
var lastSunday6 = new Date(today.setDate(today.getDate() - today.getDay() - 35));
lastSunday6 = lastSunday6.format("mm/dd/yyyy");
var lastSunday7 = new Date(today.setDate(today.getDate() - today.getDay() - 42));
lastSunday7 = lastSunday7.format("mm/dd/yyyy");
var lastSunday8 = new Date(today.setDate(today.getDate() - today.getDay() - 49));
lastSunday8 = lastSunday8.format("mm/dd/yyyy");

console.log(lastSunday8); //this gives 06/25/2017 Obviously not what I wanted.

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不需要为每个日期创建单独的变量，您可以使用for循环和数组来存储它们

var date = new Date();
date.setDate(date.getDate() - date.getDay()); //start at last sunday

var dates = [];
for (var i = 0; i < 8; i++) {
    dates.push(new Date(date));
    date.setDate(date.getDate() - 7); //subtract a week
}

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例如-以毫秒为单位计算时间增量

var today = new Date();
var lastSunday = new Date(today - today.getDay()*3600*24*1000);
console.log('Last Sunday: ' + lastSunday)
for(var i=2; i <= 8; i++) {
  lastSunday = new Date(lastSunday - 7 * 3600*24*1000);
  console.log('Sunday ' + i + ': ' + lastSunday);
}

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您每次都将日期设置为与代码行中引用的星期日相等；所以你增加了每一行，减去的天数都应该是7，试着把所有的-number改为-7，因为每一行你都在设置日期，减去7*n；其中n=周数；

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您还应该习惯使用console.log语句或cli debug逐行检查代码以检查当前持有的值变量。

为什么不简单地使用momentjs？@Andy请不要只是建议他使用库，这有点不必要。您考虑过使用for循环吗？您使用的是库吗？lastSunday.format通常会引发错误。每次执行today.setDatetoday.getDate-today.getDay-7操作时，您今天都要修改，因此只需不断减去7，并在每个操作中使用循环。并非所有白天都是24小时的夏令时，使用set和get Date方法比24小时调整时间值要好得多。这很好，但它给了我周一而不是周日。var now=新日期；var date=new Datenow.getFullYear，now.getMonth，now.getDate；加上这个，它给我星期天，谢谢@Dong将其更改为date.setUTCDatedate.getUTCDate-date.getUTCDay；我也解决了这个问题，似乎是UTC/区域设置混合的问题，导致它在UTC日结束后无法正常工作。