找到最後一個工作日給定月份PostgreSQL中

Question

找到一個給定月份的最後一個工作日在PostgreSQL的

用法：如果月底恰逢週六或週日，回到上週五，否則使用月末

例子：

• 2013年3月31日適逢星期日，所以回到2013年3月29日
• 二○一三年十一月三十○日恰逢週六，所以回到2013年11月29日

如何在PostgreSQL SQL中編寫此代碼？

我到目前爲止是這樣的（僅返回月份結束，但是當它們落在週六或週日不存在兩端月）：

SELECT as_of_dt, sum(bank_shr_bal) as bank_shr_bal 
    FROM hm_101.vw_gl_bal 
WHERE as_of_dt = (date_trunc('MONTH', as_of_dt) + INTERVAL '1 MONTH - 1 day')::date 
GROUP BY 1 

感謝

Answer 1

一種解決方案是使用CTE，發現每月數據的最後一天，每個月的實際最後一天

WITH s1 
as 
(
SELECT 
date_part('YEAR', as_of_dt) AOD_Year 
,date_part('MONTH', as_of_dt) AOD_Month 
,(date_trunc('MONTH', as_of_dt) + INTERVAL '1 MONTH - 1 day')::date AOD_MonthEnd 
,max(as_of_dt) AOD_LastFound 
    FROM hm_101.vw_gl_bal 
where (date_trunc('MONTH', as_of_dt) + INTERVAL '1 MONTH - 1 day')::date = '2013-03-31' 
group by 1, 2, 3 
) 
SELECT 
s1.AOD_MonthEnd 
,s1.AOD_LastFound 
,sum(v.bank_shr_bal) as bank_shr_bal 
    FROM hm_101.vw_gl_bal v 
INNER JOIN s1 
    on v.as_of_dt = s1.AOD_LastFound 
WHERE v.as_of_dt = '2013-03-29' 
GROUP BY 1, 2 

Answer 2

你想要做的是從本月的最後一天0-2天去除（你有）。

通過提取星期幾（DOW）並檢查它是0（星期日）還是6（星期六），我們知道要移除多少天。

你可以這樣說：

... - INTERVAL '1 day' * CASE date_part('DOW', last_day_of_month) 
          WHEN 0 THEN 2 -- Sunday, remove 2 days. 
          WHEN 6 THEN 1 -- Saturday, remove 1 day. 
          ELSE 0 -- Don't remove any days. 
         END 

對於可讀性的原因我並沒有包括在裏面完全last_day_of_month計算。

Answer 3

實際上，您可以在沒有CTE或存儲過程的情況下執行此操作。

select 
case 
    when extract(dow from last_day_of_month) = 0 
    then last_day_of_month - 2 
    when extract(dow from last_day_of_month) = 6 
    then last_day_of_month - 1 
    else 
    last_day_of_month 
end as last_weekday_of_month 
from(
    SELECT (date_trunc('MONTH', as_of_dt) 
    + INTERVAL '1 MONTH - 1 day')::date as last_day_of_month 
    from hm_101.vw_gl_bal 
)subquery; 

Answer 4

with s as (
    select *, (date_trunc('MONTH', as_of_dt) + INTERVAL '1 MONTH - 1 day')::date last_day 
    from 
    hm_101.vw_gl_bal 
) 
SELECT 
    as_of_dt, 
    gl_acct_nbr, 
    cc_nbr, 
    sum(bank_shr_bal) as bank_shr_bal 
FROM s 
WHERE as_of_dt = (
    last_day 
    - 
    (extract(dow from last_day) = 5)::int 
    - 
    2 * (extract(dow from last_day) = 6)::int 
    ) 
GROUP BY 1,2,3 

Answer 5

A calendar table真正簡化像這些查詢的SQL。 （表「平日」，實際上是基於日曆表的視圖。它的結構應該是顯而易見的。）

select max(cal_date) 
from weekdays 
where cal_date < '2013-05-01' 

或

select max(cal_date) 
from weekdays 
where cal_date between '2013-04-01' and '2013-04-30' 

Answer 6

select 
case 
    when extract(dow from first_day_of_month) = 0 then first_day_of_month 
    when extract(dow from first_day_of_month) = 1 then first_day_of_month - 1 
    when extract(dow from first_day_of_month) = 2 then first_day_of_month - 2 
    when extract(dow from first_day_of_month) = 3 then first_day_of_month - 3 
    when extract(dow from first_day_of_month) = 4 then first_day_of_month - 4 
    when extract(dow from first_day_of_month) = 5 then first_day_of_month - 5 
    when extract(dow from first_day_of_month) = 6 then first_day_of_month - 6 
    end as first_weekday_of_month, 
case 
    when extract(dow from last_day_of_month) = 6 then last_day_of_month 
    when extract(dow from last_day_of_month) = 5 then last_day_of_month - 6 
    when extract(dow from last_day_of_month) = 4 then last_day_of_month - 5 
    when extract(dow from last_day_of_month) = 3 then last_day_of_month - 4 
    when extract(dow from last_day_of_month) = 2 then last_day_of_month - 3 
    when extract(dow from last_day_of_month) = 1 then last_day_of_month - 2 
    when extract(dow from last_day_of_month) = 0 then last_day_of_month - 1 
    end as last_weekday_of_month 
from(
    SELECT 
    (date_trunc('month', current_date) -'7day'::interval)::date first_day_of_month, 
    (date_trunc('month', current_date) -'1day'::interval)::date as last_day_of_month 
)subquery;