这应该解决您的问题，请随时提供修复。该方法还将考虑给定日期的夏时制。

dateWithTimeZone = (timeZone, year, month, day, hour, minute, second) => {
  let date = new Date(Date.UTC(year, month, day, hour, minute, second));

  let utcDate = new Date(date.toLocaleString('en-US', { timeZone: "UTC" }));
  let tzDate = new Date(date.toLocaleString('en-US', { timeZone: timeZone }));
  let offset = utcDate.getTime() - tzDate.getTime();

  date.setTime( date.getTime() + offset );

  return date;
};

如何与时区和本地时间一起使用:

dateWithTimeZone("America/Los_Angeles",2019,8,8,0,0,0)

我发现最受支持的方法是使用getTimezoneOffset来计算适当的时间戳，或者更新时间，然后使用正常的方法来获得必要的日期和时间。

var mydate = new Date();
mydate.setFullYear(2013);
mydate.setMonth(02);
mydate.setDate(28);
mydate.setHours(7);
mydate.setMinutes(00);

// ET timezone offset in hours.
var timezone = -5;
// Timezone offset in minutes + the desired offset in minutes, converted to ms.
// This offset should be the same for ALL date calculations, so you should only need to calculate it once.
var offset = (mydate.getTimezoneOffset() + (timezone * 60)) * 60 * 1000;

// Use the timestamp and offset as necessary to calculate min/sec etc, i.e. for countdowns.
var timestamp = mydate.getTime() + offset,
    seconds = Math.floor(timestamp / 1000) % 60,
    minutes = Math.floor(timestamp / 1000 / 60) % 60,
    hours   = Math.floor(timestamp / 1000 / 60 / 60);

// Or update the timestamp to reflect the timezone offset.
mydate.setTime(mydate.getTime() + offset);
// Then Output dates and times using the normal methods.
var date = mydate.getDate(),
    hour = mydate.getHours();

EDIT

我以前在执行日期转换时使用UTC方法，这是不正确的。通过将偏移量添加到时间上，使用本地get函数将返回所需的结果。

我在单元测试中遇到了类似的问题(特别是开玩笑地说，单元测试在本地运行以创建快照，然后CI服务器(可能)在不同的时区运行，导致快照比较失败)。我嘲笑我们的约会和一些辅助方法，如下所示:

describe('...', () => {
  let originalDate;

  beforeEach(() => {
    originalDate = Date;
    Date = jest.fn(
      (d) => {
        let newD;
        if (d) {
          newD = (new originalDate(d));
        } else {
          newD = (new originalDate('2017-05-29T10:00:00z'));
        }
        newD.toLocaleString = () => {
          return (new originalDate(newD.valueOf())).toLocaleString("en-US", {timeZone: "America/New_York"});
        };
        newD.toLocaleDateString = () => {
          return (new originalDate(newD.valueOf())).toLocaleDateString("en-US", {timeZone: "America/New_York"});
        };
        newD.toLocaleTimeString = () => {
          return (new originalDate(newD.valueOf())).toLocaleTimeString("en-US", {timeZone: "America/New_York"});
        };
        return newD;
      }
    );
    Date.now = () => { return (Date()); };
  });

  afterEach(() => {
    Date = originalDate;
  });

});

正如马特·约翰逊所说

如果你可以限制你的使用现代网络浏览器，你现在可以做到 以下没有任何特殊库: 新的日期()。toLocaleString("en-US"， {timeZone: "America/New_York"})

这不是一个全面的解决方案，但它适用于许多场景 只需要输出转换(从UTC或本地时间到) 具体时区，但没有其他方向)。

所以尽管浏览器在创建日期时不能读取IANA时区，或者有任何方法来更改现有date对象上的时区，但似乎有一个hack围绕它:

function changeTimezone(date, ianatz) { // suppose the date is 12:00 UTC var invdate = new Date(date.toLocaleString('en-US', { timeZone: ianatz })); // then invdate will be 07:00 in Toronto // and the diff is 5 hours var diff = date.getTime() - invdate.getTime(); // so 12:00 in Toronto is 17:00 UTC return new Date(date.getTime() - diff); // needs to substract } // E.g. var here = new Date(); var there = changeTimezone(here, "America/Toronto"); console.log(`Here: ${here.toString()}\nToronto: ${there.toString()}`);

试试这样的方法，

public static getTimezoneOffset(timeZone: string, date = new Date()): number {
    const localDate = date.toLocaleString('fr', { timeZone, timeZoneName: 'long' });
    const tz = localDate.split(' ');
    const TZ = localDate.replace(tz[0], '').replace(tz[1], '').replace(' ', '');
    const dateString = date.toString();
    const offset = (Date.parse(`${dateString} UTC`) - Date.parse(`${dateString}${TZ}`)) / (3600 * 1000);
    return offset;
}

简单，支持Node.JS

传入时区与UTC时间的偏移量

function initDateInTimezone(offsetHours) {
  const timezoneOffsetInMS = offsetHours * 60 * 60000;
  let d = new Date().getTimezoneOffset() * 60000 + timezoneOffsetInMS;
  const date = new Date(new Date().getTime() - d);
    return date
}