声明给了我日期和时间。
我如何修改语句以使其仅返回日期(而不是时间)?
SELECT to_timestamp( TRUNC( CAST( epoch_ms AS bigint ) / 1000 ) );
声明给了我日期和时间。
我如何修改语句以使其仅返回日期(而不是时间)?
SELECT to_timestamp( TRUNC( CAST( epoch_ms AS bigint ) / 1000 ) );
您使用to_timestamp
函数,然后将时间戳转换为date
select to_timestamp(epoch_column)::date;
更多细节:
/* Current time */
select now(); -- returns timestamp
/* Epoch from current time;
Epoch is number of seconds since 1970-01-01 00:00:00+00 */
select extract(epoch from now());
/* Get back time from epoch */
-- Option 1 - use to_timestamp function
select to_timestamp( extract(epoch from now()));
-- Option 2 - add seconds to 'epoch'
select timestamp with time zone 'epoch'
+ extract(epoch from now()) * interval '1 second';
/* Cast timestamp to date */
-- Based on Option 1
select to_timestamp(extract(epoch from now()))::date;
-- Based on Option 2
select (timestamp with time zone 'epoch'
+ extract(epoch from now()) * interval '1 second')::date;
在你的情况下:
select to_timestamp(epoch_ms / 1000)::date;
select to_timestamp(cast(epoch_ms/1000 as bigint))::date
为我工作
在 Postgres 10 上:
SELECT to_timestamp(CAST(epoch_ms as bigint)/1000)
上面的解决方案不适用于 PostgreSQL 上的最新版本。我发现这种方法可以转换存储在数字中的纪元时间,而 int 列类型在 PostgreSQL 13 上:
SELECT TIMESTAMP 'epoch' + (<table>.field::int) * INTERVAL '1 second' as started_on from <table>;
有关更详细的说明,您可以在此处查看https://www.yodiw.com/convert-epoch-time-to-timestamp-in-postgresql/#more-214
这对我来说很好:
SELECT t.*,
to_timestamp(cast(t.prev_fire_time/1000 as bigint)) as prev_fire_time,
to_timestamp(cast(t.next_fire_time/1000 as bigint)) as next_fire_time,
to_timestamp(cast(t.start_time/1000 as bigint)) as start_time
FROM public.qrtz_triggers t;
自 GNU 纪元以来的秒数date
:
$ date +%s.%N
1627059870.945134901
这适用于 PostgreSQL 11:
# select to_timestamp (1627059870.945134901);
to_timestamp
-------------------------------
2021-07-23 19:04:30.945135+02
(1 row)
# select to_timestamp (1627059870.945134901)::date;
to_timestamp
--------------
2021-07-23
(1 row)