如何从postgresql中的时间戳中提取句点

Question

我需要从月的第一天到第15天（包括）和从第16天到月底（包括最后一天）提取和分组时间戳信息

我每周只做这个。下面是代码

SELECT 
    extract(week from dttrnreqdate) AS period, 
    SUM(bonusesCharged - bonusesCharged_rev) AS bonusesChargedForPeriod, 
    SUM(bonusesWithdrawn - bonusesWithdrawn_rev) AS bonusesWithdrawnForPeriod 
FROM 
    ls.vstatement AS v 
WHERE 
    v.dttrnreqdate >= to_timestamp('2014-12-01 00:00:00', 'yyyy-mm-dd 00:00:00') AND 
    v.dttrnreqdate < to_timestamp('2015-01-07 00:00:00', 'yyyy-mm-dd 00:00:00') 
GROUP BY 
    extract(week from dttrnreqdate) 
ORDER BY 
    extract(week from dttrnreqdate) 
ASC 

其中 dttrnreqdate是timestamp bonusesCharged, bonusesCharged_rev, bonusesWithdrawn, bonusesWithdrawn_rev - int

那么如何提取信息1-15和时间戳字段的16-30/31

Answer 1

我敢确定这回答你的问题，但我不确定它会解决你的问题。

SELECT 
    case when extract(day from dttrnreqdate) <= 15 then 'grp1' 
     else 'grp2' 
    end as grp, 
    SUM(bonusesCharged - bonusesCharged_rev) AS bonusesChargedForPeriod, 
    SUM(bonusesWithdrawn - bonusesWithdrawn_rev) AS bonusesWithdrawnForPeriod 
FROM 
    ls.vstatement AS v 
WHERE 
    v.dttrnreqdate >= to_timestamp('2014-12-01 00:00:00', 'yyyy-mm-dd 00:00:00') AND 
    v.dttrnreqdate < to_timestamp('2015-01-07 00:00:00', 'yyyy-mm-dd 00:00:00') 
GROUP BY 
    grp 
ORDER BY 
    grp 
ASC;