Question

由于我已经投了票，我决定给我的问题提供更多细节。

我的问题是我想存储次。时间仅以便于比较。对我而言，将其定义如下：

00:00:00 = (long) 0L
24:00:00 = (long) 24*60*60*1000L

PostgreSQL documentation说TIME WITHOUT TIME ZONE这个：

time [ (p) ] [ without time zone ] 
   8 bytes | time of day (no date) | 00:00:00 - 24:00:00 | 1 microsecond / 14 digits

但出于某种原因，从我的数据库到我的网络应用程序的某个地方，时间戳都搞砸了。

在下文中，我想向您展示我正在存储TIME WITHOUT TIME ZONE，从LocalTime（jodatime）映射到我的数据库中，之后将其取回。

从数据库中取回它并将其映射回LocalTime对象会给我一些像23 Feb 2016 08:00:00 GMT这样的东西。 1456214400000即> 24*60*60*1000。

有3到4个选项：

PostgresSQL实际存储整个TIMESTAMP并仅显示HH:mm:ss
jodatime正在发明这里没有的东西。
（非常不可能）我使用的映射器比我告诉他的更多。但这不太可能，因为映射器不触摸任何东西。
（可能是真的）我自己搞砸了。

更多详情：

我决定添加一些细节。我正在为我的表shop_times

private Long createShopTimes(Long shopId, DateTime dayFrom, DateTime dayTo, LocalTime timeFrom, LocalTime timeTo, DayOfWeek dayOfWeek, ShopTimesType shopTimesType) {

    Long timePeriodId = this.ctx.insertInto(SHOP_TIMES)
            .set(SHOP_TIMES.SHOP_ID, shopId)
            .set(SHOP_TIMES.DAY_OF_WEEK_ID, dayOfWeek)
            .set(SHOP_TIMES.SHOP_TIMES_TYPE_ID, shopTimesType)
            .set(SHOP_TIMES.DAY_FROM, dayFrom)
            .set(SHOP_TIMES.DAY_TO, dayTo)
            .set(SHOP_TIMES.TIME_FROM, timeFrom)
            .set(SHOP_TIMES.TIME_TO, timeTo)
            .returning(SHOP_TIMES.ID)
            .fetchOne().getValue(SHOP_TIMES.ID);

    List<ShopTimesRecord> fetchInto = this.ctx.select(
            SHOP_TIMES.TIME_FROM,
            SHOP_TIMES.TIME_TO
            )
            .from(SHOP_TIMES)
            .fetchInto(ShopTimesRecord.class);

    for (ShopTimesRecord shopTimesRecord : fetchInto) {

        if(shopTimesRecord.getTimeFrom().toDateTimeToday().getMillis() > 24*60*60*1000L) {
            System.err.println("This should not happen..");
        }

        Date from = new Date(shopTimesRecord.getTimeFrom().toDateTimeToday().getMillis());
        Date to = new Date(shopTimesRecord.getTimeTo().toDateTimeToday().getMillis());

        System.out.println(from.toGMTString());
        System.out.println(shopTimesRecord.getTimeFrom().toDateTimeToday().getMillis());

        System.out.println(to.toGMTString());
        System.out.println(shopTimesRecord.getTimeTo().toDateTimeToday().getMillis());
    }

    return timePeriodId;
}

正如你所看到的，我现在得到了一些我不会想到的东西：

This should not happen..
23 Feb 2016 08:00:00 GMT
1456214400000
23 Feb 2016 20:00:00 GMT
1456257600000

这是我用来存储时间信息的表shop_times：

CREATE TABLE shop_times (

    -- PRIMARY KEY

    id BIGSERIAL PRIMARY KEY,

    -- FOREIGN KEYS

    shop_id BIGINT NOT NULL,

    CONSTRAINT fk__shop_times__shop
        FOREIGN KEY (shop_id)
        REFERENCES shop(id),

    shop_times_type_id BIGINT NOT NULL,

    CONSTRAINT fk_shop_times__shop_times_type
        FOREIGN KEY (shop_times_type_id)
        REFERENCES shop_times_type(id),

    day_of_week_id BIGINT NOT NULL,

    CONSTRAINT fk__shop_times__day_of_week
        FOREIGN KEY (day_of_week_id)
        REFERENCES day_of_week(id),

    -- ATTRIBUTES

    day_from TIMESTAMP WITH TIME ZONE NOT NULL, 
    day_to TIMESTAMP WITH TIME ZONE NOT NULL,

    time_from TIME WITHOUT TIME ZONE NOT NULL,
    time_to TIME WITHOUT TIME ZONE NOT NULL,

    -- CONSTRAINTS

    CHECK(day_from < day_to),

    CHECK(time_from < time_to)

);

我正在使用的映射器。但正如你所看到的那样，它只是花费时间并将其传递出去;

public class TimeWithoutTzToJodaLocalTimeConverter implements Converter<Time, LocalTime> {
    private static final long serialVersionUID = -2736422625956918206L;

    @Override
    public LocalTime from(Time timestamp) {
        LocalTime dateTime = new LocalTime(timestamp.getTime());
        return dateTime;
    }

    @Override
    public Time to(LocalTime localTime) {
        Time time = new Time(localTime.toDateTimeToday().getMillis());
        return time;
    }

    @Override
    public Class<Time> fromType() {
        return Time.class;
    }

    @Override
    public Class<LocalTime> toType() {
        return LocalTime.class;
    }

}

Answer 1

抱歉 - 不要相信你。

=> SELECT '24:00:00'::time without time zone;
   time   
----------
 24:00:00
(1 row)

=> SELECT '24:00:01'::time without time zone;
ERROR:  date/time field value out of range: "24:00:01"
LINE 1: SELECT '24:00:01'::time without time zone;

Answer 2

您正在创建完整的DateTime：

.toDateTimeToday()

http://joda-time.sourceforge.net/api-release/org/joda/time/TimeOfDay.html#toDateTimeToday%28%29

我的时间戳在哪里搞砸了＃34;？

2 个答案: