java - 不支持的字段 : Year when formatting an instant to Date ISO

Question

这个问题在这里已经有了答案:

UnsupportedTemporalTypeException when formatting Instant to String (7 个回答)

关闭 6年前。

我正在尝试将 Instant 格式化为 ldap 日期 ISO8601，但在 f.format(Instant.now()); 处失败:

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));