Redshift的时差

Ben*_*nny 9 sql date-arithmetic amazon-redshift

如何获得两列之间的确切时间差异

例如:

   col1 date is 2014-09-21 02:00:00
   col2 date is 2014-09-22 01:00:00
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输出像

result: 23:00:00
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我得到了结果

 Hours Minutes Seconds
 --------------------
  3    3       20
  1    2       30
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使用以下查询

SELECT start_time,
       end_time,
       DATE_PART(H,end_time) - DATE_PART(H,start_time) AS Hours,
       DATE_PART(M,end_time) - DATE_PART(M,start_time) AS Minutes,
       DATE_PART(S,end_time) - DATE_PART(S,start_time) AS Seconds
FROM user_session
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但我需要

 Difference 
 -----------
  03:03:20
  01:02:30
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Joe*_*ris 14

使用DATEDIFF获取两个日期时间之间的秒数:

DATEDIFF(second,'2014-09-23 00:00:00.000','2014-09-23 01:23:45.000')
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然后使用DATEADD将秒添加到'1900-01-01 00:00:00':

DATEADD(seconds,5025,'1900-01-01 00:00:00')
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然后将结果CAST转换为TIME数据类型(请注意,这会将您限制为最多 24小时):

CAST('1900-01-01 01:23:45' as TIME)
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然后LTRIM将日期部分值关闭TIME数据(由Benny发现).Redshift不允许在实际存储的数据上使用TIME :

LTRIM('1900-01-01 01:23:45','1900-01-01')
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现在,只需一步即可完成:

SELECT LTRIM(DATEADD(seconds,DATEDIFF(second,'2014-09-23 00:00:00','2014-09-23 01:23:45.000'),'1900-01-01 00:00:00'),'1900-01-01');
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:)