Java8 解析给定字符串的日期或日期时间格式

Question

我有一个文件可以有日期修改值，格式为日期或日期时间。我曾经将值解析为：

String t = "2012-01-05T21:21:52.834Z";
logger.info(ZonedDateTime.parse(t).toEpochSecond() * 1000);

现在，字符串也可以是

t = "2012-01-05";

引发错误

Exception in thread "main" java.time.format.DateTimeParseException: Text '2012-01-05' could not be parsed at index 10 at java.time.format.DateTimeFormatter.parseResolved0(Unknown Source)

如果我用 (Reference)

做这个字符串

LocalDate date = LocalDate.parse(t, DateTimeFormatter.ISO_DATE);
logger.info(date.atStartOfDay(ZoneId.of("UTC")).toEpochSecond() * 1000);

这行得通。但是，正如我所提到的，字符串可能是这些类型中的任何一种，我如何识别格式然后相应地获取毫秒数？

Answer 1

一种可能的解决方案是使用具有默认值的可选模式。使用 DateTimeFormatterBuilder, you can append the wanted pattern with the time part in an optional section, i.e. surrounded by [...]. In the case where the fields are absent, we provide default values by setting them to 0. The OFFSET_SECONDS 字段为 0 表示与 UTC 没有偏移量。

public static void main(String[] args) {
    String[] dates = { "2012-01-05T21:21:52.834Z", "2012-01-05" };

    DateTimeFormatter formatter = 
        new DateTimeFormatterBuilder().appendPattern("yyyy-MM-dd['T'HH:mm:ss.SSSz]")
                                      .parseDefaulting(ChronoField.HOUR_OF_DAY, 0)
                                      .parseDefaulting(ChronoField.MINUTE_OF_HOUR, 0)
                                      .parseDefaulting(ChronoField.SECOND_OF_MINUTE, 0)
                                      .parseDefaulting(ChronoField.NANO_OF_SECOND, 0)
                                      .parseDefaulting(ChronoField.OFFSET_SECONDS, 0)
                                      .toFormatter();
    for (String date : dates) {
        ZonedDateTime zonedDateTime = ZonedDateTime.parse(date, formatter);
        System.out.println(zonedDateTime.toEpochSecond() * 1000);
    }
}

Java8 解析给定字符串的日期或日期时间格式

Java8 parse date or date time format for a given string

java

date

java-8

java-time