unix_timestamp与转换为时间戳之间的差异

Question

我有一个配置单元表的情况，要将数字字符串的两个字段（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。我尝试了两种方法：

• 方法1：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

并且将输出作为

• 方法2：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;

并获得输出为

我不清楚为什么输出会有所不同。

Answer 1

方法2最终等于这样的东西

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

您不能像这样减去日期。必须使用DateDiff。

仅当存在差异时，Hive中的DateDiff返回> 0，否则您将得到零。

Answer 2

unix_timestamp（）给您纪元时间，即。自UNIX时代以来的时间（以秒为单位）1970-01-01 00:00:00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（），减去后再使用from_unixtime（）返回例如。

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