我有以下格式的日期: 2010-03-01T00:00:00-08:00我向它拋出了以下SimpleDateFormats來對其進(jìn)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 };我有一個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;}它似乎已擊中該模式,yyyyMMddHHmm但將日期返回為Thu Dec 03 00:01:00 PST 2009。解析此日期的正確模式是什么?更新:我不需要時區(qū)解析。我不希望在區(qū)域之間移動對時間敏感的問題,但是我將如何解析“ -08:00”區(qū)域格式?單元測試:@Testpublic 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"));}
Java SimpleDateFormat的時區(qū)帶有冒號分隔符?
慕碼人8056858
2019-11-13 13:08:28