看起来像以毫秒为单位的unix / epoch时间?
declare @x bigint = 1392979784822
declare @msin1day bigint = 3600 * 24 * 1000
select dateadd(ms, @x % @msin1day, dateadd(day, @x / @msin1day, '19700101'))
(No column name)
2014-02-21 10:49:44.823
如何在SQL中将GPS格式的时间转换为本地时间格式?
看起来像以毫秒为单位的unix / epoch时间?
declare @x bigint = 1392979784822
declare @msin1day bigint = 3600 * 24 * 1000
select dateadd(ms, @x % @msin1day, dateadd(day, @x / @msin1day, '19700101'))
(No column name)
2014-02-21 10:49:44.823