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

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Format Instant to String

(7个答案)


4年前关闭。




我正在尝试将Instant格式化为ldap日期ISO8601，但它在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)
...
...

最佳答案

要格式化Instant，需要一个时区。

 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));

关于java - 不支持的字段: Year when formatting an instant to Date ISO [duplicate]，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/40211892/