我有以下格式的日期: 2010-03-01T00:00:00-08:00
我向它抛出了以下SimpleDateFormats来对其进行解析:
private static final SimpleDateFormat[] FORMATS = { new SimpleDateFormat("yyyy-MM-dd'T'HH:mm:ssZ"), //ISO8601 long RFC822 zone new SimpleDateFormat("yyyy-MM-dd'T'HH:mm:ssz"), //ISO8601 long long form zone new SimpleDateFormat("yyyy-MM-dd'T'HH:mm:ss"), //ignore timezone new SimpleDateFormat("yyyyMMddHHmmssZ"), //ISO8601 short new SimpleDateFormat("yyyyMMddHHmm"), new SimpleDateFormat("yyyyMMdd"), //birthdate from NIST IHE C32 sample new SimpleDateFormat("yyyyMM"), new SimpleDateFormat("yyyy") //just the year };
我有一个使用如下格式的便捷方法:
public static Date figureOutTheDamnDate(String wtf) { if (wtf == null) { return null; } Date retval = null; for (SimpleDateFormat sdf : FORMATS) { try { sdf.setLenient(false) retval = sdf.parse(wtf); System.out.println("Date:" + wtf + " hit on pattern:" + sdf.toPattern()); break; } catch (ParseException ex) { retval = null; continue; } } return retval; }
它似乎已击中该模式,yyyyMMddHHmm但将日期返回为Thu Dec 03 00:01:00 PST 2009。
解析此日期的正确模式是什么?
更新:我不需要时区解析。我不希望在区域之间移动对时间敏感的问题,但是如何解析“ -08:00”区域格式???
单元测试:
@Test public void test_date_parser() { System.out.println("\ntest_date_parser"); //month is zero based, are you effing kidding me Calendar d = new GregorianCalendar(2000, 3, 6, 13, 00, 00); assertEquals(d.getTime(), MyClass.figureOutTheDamnDate("200004061300")); assertEquals(new GregorianCalendar(1950, 0, 1).getTime(), MyClass.figureOutTheDamnDate("1950")); assertEquals(new GregorianCalendar(1997, 0, 1).getTime(), MyClass.figureOutTheDamnDate("199701")); assertEquals(new GregorianCalendar(2010, 1, 25, 15, 19, 44).getTime(), MyClass.figureOutTheDamnDate("20100225151944-0800")); //my machine happens to be in GMT-0800 assertEquals(new GregorianCalendar(2010, 1, 15, 13, 15, 00).getTime(),MyClass.figureOutTheDamnDate("2010-02-15T13:15:00-05:00")); assertEquals(new GregorianCalendar(2010, 1, 15, 18, 15, 00).getTime(), MyClass.figureOutTheDamnDate("2010-02-15T18:15:00-05:00")); assertEquals(new GregorianCalendar(2010, 2, 1).getTime(), MyClass.figureOutTheDamnDate("2010-03-01T00:00:00-08:00")); assertEquals(new GregorianCalendar(2010, 2, 1, 17, 0, 0).getTime(), MyClass.figureOutTheDamnDate("2010-03-01T17:00:00-05:00")); }
单元测试的输出:
test_date_parser Date:200004061300 hit on pattern:yyyyMMddHHmm Date:1950 hit on pattern:yyyy Date:199701 hit on pattern:yyyyMM Date:20100225151944-0800 hit on pattern:yyyyMMddHHmmssZ Date:2010-02-15T13:15:00-05:00 hit on pattern:yyyy-MM-dd'T'HH:mm:ss Date:2010-02-15T18:15:00-05:00 hit on pattern:yyyy-MM-dd'T'HH:mm:ss Date:2010-03-01T00:00:00-08:00 hit on pattern:yyyy-MM-dd'T'HH:mm:ss Date:2010-03-01T17:00:00-05:00 hit on pattern:yyyy-MM-dd'T'HH:mm:ss
JodaTime的DateTimeFormat救援:
DateTimeFormat
String dateString = "2010-03-01T00:00:00-08:00"; String pattern = "yyyy-MM-dd'T'HH:mm:ssZ"; DateTimeFormatter dtf = DateTimeFormat.forPattern(pattern); DateTime dateTime = dtf.parseDateTime(dateString); System.out.println(dateTime); // 2010-03-01T04:00:00.000-04:00
(时区和时区的差异toString()仅是因为我在GMT-4,并且没有明确设置语言环境)
如果要结束java.util.Date使用DateTime#toDate():
java.util.Date
DateTime#toDate()
Date date = dateTime.toDate();
等待JDK7(JSR-310) JSR-310,如果你想在标准Java SE API中使用更好的格式化程序,则引用实现称为ThreeTen(希望它将实现到Java 8中)。SimpleDateFormat在时区表示法中,当前确实不会吃掉冒号。
SimpleDateFormat
更新:根据更新,你显然不需要时区。这应该与SimpleDateFormat。只需Z在模式中忽略它。
String dateString = "2010-03-01T00:00:00-08:00"; String pattern = "yyyy-MM-dd'T'HH:mm:ss"; SimpleDateFormat sdf = new SimpleDateFormat(pattern); Date date = sdf.parse(dateString); System.out.println(date); // Mon Mar 01 00:00:00 BOT 2010
(根据我的时区是正确的)