从星期一开始的周间隔查询

debugcn 发表于 Dev

莫恩

我需要做一个JasperReport。我需要显示账户过程的总数，分解成weekly intervals与number中activated和declined账户。

到目前为止，对于每周间隔查询：

SELECT *
FROM account_details
WHERE DATE date_opened = DATE_ADD(2014-01-01, INTERVAL(1-DAYOFWEEK(2014-01-01)) +1 DAY)

这似乎是正确的，但不POSTGRES正确。它一直在抱怨1-DAYOFWEEK。这是我有望实现的目标：

在此处输入图片说明

更新这是非常丑陋的，但我不知道有什么更好的。Id可以完成工作。但是不知道它是否可以重构为至少看起来更好。我现在也不知道如何处理被零除的问题。

        SELECT to_char(d.day, 'YYYY/MM/DD  -  ') || to_char(d.day + 6, 'YYYY/MM/DD') AS Month
         , SUM(CASE WHEN LOWER(situation) LIKE '%active%' THEN 1 ELSE 0 END)     AS Activated
         , SUM(CASE WHEN LOWER(situation) LIKE '%declined%' THEN 1 ELSE 0 END) AS Declined
         , SUM(CASE WHEN LOWER(situation) LIKE '%declined%' OR LOWER(situation) LIKE '%active%' THEN 1 ELSE 0 END) AS Total
         , to_char( 100.0 *( (SUM(CASE WHEN LOWER(situation) LIKE '%active%' THEN 1 ELSE 0 END)) / (SUM(CASE WHEN LOWER(situation) LIKE '%declined%' OR LOWER(situation) LIKE '%active%' THEN 1 ELSE 0 END))::real) , '99.9') AS percent_activated
         , to_char( 100.0 *( (SUM(CASE WHEN LOWER(situation) LIKE '%declined%' THEN 1 ELSE 0 END)) / (SUM(CASE WHEN LOWER(situation) LIKE '%declined%' OR LOWER(situation) LIKE '%active%' THEN 1 ELSE 0 END))::real) , '99.9') AS percent_declined
    FROM   (
       SELECT day::date
       FROM   generate_series('2014-08-01'::date, '2014-09-14'::date, interval '1 week') day
       ) d
    JOIN   account_details a ON a.date_opened >= d.day 
                            AND a.date_opened <  d.day + 6
    GROUP  BY d.day;

欧文·布兰德斯特（Erwin Brandstetter）

SELECT to_char(d.day, 'YYYY/MM/DD" - "')
    || to_char(d.day + 6, 'YYYY/MM/DD') AS week
     , count(situation ILIKE '%active%' OR NULL) AS activated
     , ...
FROM   (
   SELECT day::date
   FROM   generate_series('2014-08-11'::date
                        , '2014-09-14'::date
                        , '1 week'::interval) day
   ) d
LEFT   JOIN account_details a ON a.date_opened >= d.day 
                             AND a.date_opened <  d.day + 7  -- 7, not 6!
GROUP  BY d.day;