Question

我有一个从不同时区使用的应用程序。如果有人进入纽约（东部时间）的数据库，它应该显示在纽约时间输入的人（它通过大部分应用程序完成）。但是，当来自时区之外的某人查看它时，似乎时间被转换为该用户的时区。

我正在尝试停止转换或将转换转换为正确的时间。它正确传递给javascript，它位于发现问题的javascript中。

对象的日期字符串显示（例如）“Fri Aug 21 12:30:00 PDT 2013”。

我修改了this function以使用美国夏令时并使用中央时区。我还使用了this function来处理从格式到unix时代的日期。

我在想，如果“PDT”或任何其他时区标识符（“CST”，“MST”，“PST”等），则可能不会发生转换。我会注意到除了“PDT”之外的时间戳是我需要的正确时间，但有些事情会在2小时前显示出来。

我的主要问题是，如果有人在2013年7月24日下午1点在纽约进入，那么无论我在查看数据时处于哪个时区，都应该说。这是问题所在：

用户从一个位置进入系统（并非总是如此 2013年7月24日，纽约，但它很容易作为一个例子使用下午3:00。
从数据库和应用程序发送正确的javascript（2013年7月24日下午3:00）
如果我处于不同的时区（让我们说中央），那么日期是在javascript中的对象中显示为（Wed Jul 24 3:00 CDT 2013）。＆lt; - 正如你所看到的那样在技术上仍然是正确的 “CDT”的时区标识符。如果时间基本相同，如果从某种程度上来说，删除它的标识符，我想我会是金色的。
未调用任何其他功能，现在显示日期为7/24/2013 2:00。

任何想法都会非常有帮助。这是我的转换功能：

function calcTime(date) {
//This is to pad the date with a zero if the
//number is less than 10. eg 7/1/2013 = 07/01/2013
function pad(num) {
    num = num.toString();
    if (num.length == 1) return "0" + num;
    return num;
}

//I needed to have my date passed in as string
//This will set the date to the string and then get the time from
//that date. It will then set that time.
var date = new Date(date);
var time = date.getTime();
date.setTime(time);

var year = date.getUTCFullYear();

//This nifty little thing will get you the daylight savings time
// and adjust correctly. In the US, prior to 2006 the DST started
//first sunday in April and ended in the last sunday in October.
//After 2007, it was changed to start on the 2nd sunday in March
//and end in the first sunday in November. If you are working
//with a different country, please follow this link for
//the equations in relation to their daylight savings time.
//http://www.webexhibits.org/daylightsaving/i.html
if (year <= 2006) {
    var start_day = (2 + ((((6 * year) - year / 4) % 7) + 1));
    var DST_start = new Date(Date.UTC(year, 3, start_day, 1, 0, 0));

    var end_day = (31 - ((((5 * year) / 4) + 1) % 7));
    var DST_end = new Date(Date.UTC(year, 9, end_day, 1, 0, 0));
}

if (year >= 2007) {
    start_day = (14 - ((1 + (year * 5) / 4) % 7));
    DST_start = new Date(Date.UTC(year, 2, start_day, 1, 0, 0));

    end_day = (7 - ((1 + (year * 5) / 4) % 7));
    DST_end = new Date(Date.UTC(year, 10, end_day, 1, 0, 0));
}

//This function is supposed to make sure no matter where you are,
//you can see this in US central time. Obviously you will need to
//change this if you need a different time zone.
//All timezones to the west of GMT to the International Date Line will be negative.
//Timezones east of GMT to the International Date Line will be positive.
//The below offset is if it is DST, it adjusts the time accordingly.
var centralOffset = -6 * 60 * 60 * 1000;
if (date > DST_start && date < DST_end) centralOffset = -5 * 60 * 60 * 1000;

date.setTime(time + centralOffset);

//This will call the pad function to return a two digit month/day format.
//Dates are zero based in JS and needed to add 1 to the UTCMonth so we receive the
//current month
var centralTime = pad(date.getUTCMonth() + 1) + "/" +
pad(date.getUTCDate()) + "/" + date.getUTCFullYear() +
" " + pad(date.getUTCHours()) + ":" +
pad(date.getUTCMinutes()) + ":" + pad(date.getUTCSeconds());

return centralTime;
};

//Test Data
var date = "7/25/2013 09:30:00";
calcTime(date);

更新我运气不好。我已经能够成功删除时区标识符，但是当我有一个正确日期的新字符串并尝试将其传递给新的Date函数时，它会重新放入标识符。请参阅下面的更新代码。

function getDateTime() {
  function pad(num) {
    num = num.toString();
    if (num.length == 1) return "0" + num;
    return num;
}


var timezones = ["PST", "PDT", "MST", "MDT", "CST", "CDT", "EST", "EDT", "GMT"];

var date = this; //this = "Wed Aug 21 14:38:00 CST 2013" or whatever date is being passed in

//This converts date to string so it can remove the time zone identifier.
date = date.toDateString();

var length = timezones.length;
while (length--) {
    if (date.indexOf(timezones[length]) != -1) {
        date = date.replace(timezones[length], "");
    }
}

date = new Date(date); // This pretty much defeats all that I did above.
//date = Date(date); //This will give the current date and time.

var month = pad(date.getMonth() + 1);
var day = pad(date.getDate());
var year = date.getFullYear();
var hour = pad(date.getHours());
var minute = pad(date.getMinutes());
var second = pad(date.getSeconds());


if (hour > 12) {
    var timestring = hour.toString() + ":" + minute.toString() + ":" + second.toString() + " PM";
} else {
    timestring = hour.toString() + ":" + minute.toString() + ":" + second.toString() + " AM";
}

var output = month.toString() + "/" + day.toString() + "/" + year.toString() + " " +   timestring;

return output;

}

希望这可以清除其他任何人在原帖中可能产生的混淆。

Answer 1

使用多个字符串替换来保留原始字符串：

var foo = "2013-10-01T09:00:00.000-04:00";
var bar = foo.substr(-6);
var baz = foo.substr(11,2)
var isoDate = new Date(foo).toISOString().replace("Z",bar).replace(/T../,"T"+baz);
var hours = String(baz).concat(Number(baz) < 12 ? "PM" : "AM").replace(/^0/,"")

防止JavaScript中的隐式日期转换

1 个答案: