Question

我正在处理一个查询，该查询生成按created和type分组的大小列的运行总和，并且还会考虑deleted日期。所以我并不仅仅意味着该日期的总和，而是一笔运行金额。如果我周一把一个苹果放在桶里（星期一= 1）。如果我在星期二再放一个苹果（星期二= 2），... 我只是对如何开始感到难过。

以下是一些示例数据：

[id,type, size, created, deleted]  
1,A,    2,  2014-08-05, 2014-08-06  
2,A,    3,  2014-08-05,   
3,A,    5,  2014-08-06,       
4,A,    4,  2014-08-06,      
5,B,    2,  2014-08-06, 2014-08-06  
6,B,    4,  2014-08-07, 2014-08-07  
7,C,    6,  2014-08-07, 2014-08-07  
8,C,    4,  2014-08-07,       
9,D,    3,  2014-08-07,       
10,E,   5,  2014-08-07,       
11,C,   6,  2014-08-07,

结果应如下所示：

[Date, Type, Sum]  
2014-08-05,A,5  (sum of IDs 1,2)  
2014-08-06,A,14 (sum of IDs 1,2,3,4)  
2014-08-06,B,2  
2014-08-07,A,11 (sum of IDs 2,3,4: Notice that ID 1 is not included because it was deleted 2014-08-06)  
2014-08-07,B,6 (sum of IDs 5,6)  
2014-08-07,C,16 (sum of IDs 7,8,11)  
etc...

Answer 1

SELECT DISTINCT day, type
     , sum(size) OVER (PARTITION BY type ORDER BY day) AS running_total
FROM (
   SELECT created AS day, type, size
   FROM   tbl

   UNION  ALL
   SELECT deleted, type, size * -1
   FROM   tbl
   WHERE  deleted IS NOT NULL
   ) sub
ORDER BY 1,2;

返回：

day         type  running_total
-------------------------------
2014-08-05  A     5
2014-08-06  A     12
2014-08-06  B     0
2014-08-07  B     0
2014-08-07  C     10
2014-08-07  D     3
2014-08-07  E     5

说明：

按照(created, type)给出的每一行和数字。
对已删除该行的(deleted, type)执行相同的操作。
使用窗口聚合函数sum() OVER ...
由于每天可以生成多个相同的行，因此请DISTINCT应用折叠。

如果删除仅在第二天生效，只需在deleted日期添加1：

...
   SELECT deleted + 1, type, size * -1
...

更快

结果相同，更详细，但更快：

SELECT day, type
     , sum(daily_delta) OVER (PARTITION BY type ORDER BY day) AS running_total
FROM  (
    SELECT day, type, sum(size) AS daily_delta
    FROM (
        SELECT created AS day, type, size
        FROM   tbl

        UNION  ALL
        SELECT deleted, type, size * -1
        FROM   tbl
        WHERE  deleted IS NOT NULL
        ) sub
    GROUP  BY  1,2
    ) sub
ORDER BY 1,2;

与详细解释：

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PostgreSQL: running count of rows for a query 'by minute'

SQL Fiddle.

棘手的SQL组由sql查询与运行总和

1 个答案:

说明：

更快