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我有一个显示来自 XML 文件的数据的谷歌地图应用程序。当您将鼠标悬停在某物上时，它会返回一些信息，包括数据。XML文件中的日期是这样的......

  <cap:expires>2013-05-02T00:00:00-05:00</cap:expires>

我有这段代码可以转换它....

function dateFromString(s) {

  var bits = s.split(/[-T:+]/g);
  var d = new Date(bits[0], bits[1]-1, bits[2]);
  d.setHours(bits[3], bits[4], bits[5]);

  // Get supplied time zone offset in minutes
  var offsetMinutes = bits[6] * 60 + Number(bits[7]);
  var sign = /\d\d-\d\d:\d\d$/.test(s)? '-' : '+';

  // Apply the sign
  offsetMinutes = 0 + (sign == '-'? -1 * offsetMinutes : offsetMinutes);

  // Apply offset and local timezone
  d.setMinutes(d.getMinutes() - offsetMinutes - d.getTimezoneOffset())

  // d is now a local time equivalent to the supplied time
return (d);

} 

var days = ["Sunday","Monday","Tuesday","Wednesday","Thursday","Friday","Saturday"];
var months =['January','February','March','April','May','June','July','August','September','October','November','December'];
var ampm = " am";

var dt = (dateFromString("yyyy-MM-dd'T'HH:mm:ssZ"));
var yr = dt.getFullYear();
var mth = dt.getMonth();  // months in Javascript are 0-11 so May is month 4
mth = months[mth];
var dte = dt.getDate();
var dy = dt.getDay();  // days are 0-6
dy = days[dy];
var hrs = dt.getHours();
var h1 = hrs;
var mins = dt.getMinutes();

if (hrs >= 12) {ampm = " pm"}
if (hrs >= 13) {hrs = hrs - 12}
if (h1 == 0) {hrs = 12}

if (hrs <10) {hrs = "0" + hrs};  // if  leading zero desired
if (mins <10) {mins = "0" + mins};

var dtstring = dy + " " + mth + " " + dte + " " + yr + " " + hrs + ":" + mins + ampm;

当您对其进行硬编码时，它会很好地转换它。

var dt = (dateFromString( '2013-05-02T11:08:00-6:00'));

我的问题是如何以及在何处插入 XML 中的元素，以便它知道要转换什么？我将它设置为输出，我只是不确定将输入放在哪里进行转换。我已经包含了指向地图和 xml 文件的链接，以防有人需要查看完整代码以了解它是如何设置的。

演示图

演示 XML

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1 回答 1

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Java中有一个特殊的类来处理XML dateTime javax.xml.datatype.XMLGregorianCalendar

XMLGregorianCalendar xgc = DatatypeFactory.newInstance().newXMLGregorianCalendar("2013-05-02T00:00:00-05:00");

现在使用它的方法来获取必要的字段。您还可以将其转换为 GregorianCalendar 和 Date

于 2013-05-03T05:08:18.063 回答