unix_timestamp 和强制转换为时间戳之间的区别

Question

我有一个配置单元表的情况，将两个数字字符串字段（T1 和 T2）转换为日期时间戳格式“YYYY-MM-DD hh:mm:ss.SSS”并找出两者的区别. 我试过两种方法：

• 方法一：通过CAST

Select CAST(regexp_replace(substring(t1, 1,17),'(\\d{4})(\\d{2})(\\d{2})(\\d{2})(\\d{2})(\\d{2})(\\d{3})','$1-$2-$3 $4:$5:$6.$7') as timestamp), CAST(regexp_replace(substring(t2, 1,17),'(\\d{4})(\\d{2})(\\d{2})(\\d{2})(\\d{2})(\\d{2})(\\d{3})','$1-$2-$3 $4:$5:$6.$7') as timestamp), CAST(regexp_replace(substring(t1, 1,17),'(\\d{4})(\\d{2})(\\d{2})(\\d{2})(\\d{2})(\\d{2})(\\d{3})','$1-$2-$3 $4:$5:$6.$7') as timestamp) - CAST(regexp_replace(substring(t2, 1,17),'(\\d{4})(\\d{2})(\\d{2})(\\d{2})(\\d{2})(\\d{2})(\\d{3})','$1-$2-$3 $4:$5:$6.$7') as timestamp) as time_diff
from tab1

得到输出

• 方法二：通过unix_timestamp

Select from_unixtime (unix_timestamp(substring(t1,1,17),'yyyyMMddhhmmssSSS'),'yyyy-MM-dd hh:mm:ss.SSS'), from_unixtime (unix_timestamp(substring(t2,1,17),'yyyyMMddhhmmssSSS'),'yyyy-MM-dd hh:mm:ss.SSS'), from_unixtime (unix_timestamp(substring(t1,1,17),'yyyyMMddhhmmssSSS'),'yyyy-MM-dd hh:mm:ss.SSS') - from_unixtime (unix_timestamp(substring(t2,1,17),'yyyyMMddhhmmssSSS'),'yyyy-MM-dd hh:mm:ss.SSS') as time_diff
from tab1;

得到输出

我不清楚为什么输出会有差异。

Answer 1

unix_timestamp() 为您提供纪元时间，即。自 unix epoch 1970-01-01 00:00:00 以来的时间（以秒为单位） 而时间戳将提供日期和时间，即 YYYY-MM-DD T HH:MI:SS 因此，一种准确的方法是将字符串时间戳转换为 unix_timestamp()，减去然后使用 from_unixtime() 转换回来 例如。

select from_unixtime(unix_timestamp('2020-04-12 01:30:02.000') - unix_timestamp('2020-04-12 01:29:43.000'))

Answer 2

方法2最终等同于这样的东西

select ('2020-04-12 01:30:02.000' - '2020-04-12 01:29:43.000') as time_diff;

你不能像这样减去日期。你必须使用 DateDiff。

在 Hive 中 DateDiff 仅在当天有差异时返回 > 0，否则您将得到零。