不支持的字段：将即时格式化为日期ISO的年份

我试图将Instant格式化为ldap日期ISO8601，但它在f.format（Instant.now（））失败; ：

String input = "20161012235959.0Z"; DateTimeFormatter f = DateTimeFormatter.ofPattern ( "uuuuMMddHHmmss[,S][.S]X" ); OffsetDateTime odt = OffsetDateTime.parse ( input , f ); Instant instant = odt.toInstant (); f.format(Instant.now()); 

错误是：

 java.time.temporal.UnsupportedTemporalTypeException: Unsupported field: Year at java.time.Instant.getLong(Instant.java:603) at java.time.format.DateTimePrintContext.getValue(DateTimePrintContext.java:298) at java.time.format.DateTimeFormatterBuilder$NumberPrinterParser.format(DateTimeFormatterBuilder.java:2540) at java.time.format.DateTimeFormatterBuilder$CompositePrinterParser.format(DateTimeFormatterBuilder.java:2179) at java.time.format.DateTimeFormatter.formatTo(DateTimeFormatter.java:1746) at java.time.format.DateTimeFormatter.format(DateTimeFormatter.java:1720) ... ... 

Answer 1

要格式化Instant，需要时区。

  String input = "20161012235959.0Z"; DateTimeFormatter f = DateTimeFormatter .ofPattern ( "uuuuMMddHHmmss.SX" ) .withLocale( Locale.FRANCE ) .withZone( ZoneId.of("UTC")); OffsetDateTime odt = OffsetDateTime.parse ( input , f ); Instant instant = odt.toInstant (); System.out.println(input); System.out.print(f.format(instant));