Question

我有以下数据表：

month      marketid   totalsold   totalshipped   lefttoship_thismonth
....
01-01-2015   1          100            50              50
01-01-2015   2          10             3               7
01-01-2015   3          0              0               0
01-02-2015   1          0              50             -50
01-02-2015   2          20             0               20
01-02-2015   3          0              0                0

此表基本上显示了每月每个市场的订单和发货量信息。 01-01-2015列中的month日期实际上代表Jan 2015（整个月）。

我希望每月每个市场SUM lefttoship_thismonth与前几个月相比。这是必要的，因为有人可以在2月份提供1月订单。所以我想知道每个月还需要运送多少物品。

输出应为：

month      marketid   totalsold   totalshipped   totallefttoship   TOTALLEFT
01-01-2015   1          100            50              50             50   
01-01-2015   2          10             3               7               7
01-01-2015   3          0              0               0              0
01-02-2015   1          0              50             -50             0    /50-50
01-02-2015   2          20             0               20             27  /7+20
01-02-2015   3          0              0                0               0  / 0+0

我该怎么做？我不知道如何以这种方式加总，month列非常难以使用。

Answer 1

Sum()Over()窗口聚合函数

SELECT "month",
       marketid,
       totalsold,
       totalshipped,
       lefttoship_thismonth,
       Sum(lefttoship_thismonth)OVER(partition BY marketid ORDER BY month ) AS TOTALLEFT
FROM   yourtable
ORDER  BY "month",
          marketid

Live Demo

Answer 2

如果您的PostgreSQL版本（还）不允许窗口函数，您可以使用SubQuery执行此操作：

WITH t (month, marketid, totalsold, totalshipped, lefttoship_thismonth) AS
(VALUES
    ('01-01-2015'::date,   1, 100, 50,  50),
    ('01-01-2015'::date,   2,  10,  3,   7),
    ('01-01-2015'::date,   3,   0,  0,   0),
    ('01-02-2015'::date,   1,   0, 50, -50),
    ('01-02-2015'::date,   2,  20,  0,  20),
    ('01-02-2015'::date,   3,   0,  0,   0) 
)

SELECT
    month, 
    marketid, 
    totalsold, 
    totalshipped, 
    lefttoship_thismonth, 
    (SELECT sum(lefttoship_thismonth) 
       FROM t t2 
      WHERE t2.marketid  = t1.marketid AND
            t2.month    <= t1.month
    ) AS total_left
FROM 
    t t1
ORDER BY
    month, marketid ;

会得到以下结果：

|------------+----------+-----------+--------------+----------------------+------------|
|   month    | marketid | totalsold | totalshipped | lefttoship_thismonth | total_left |
|------------+----------+-----------+--------------+----------------------+------------|
| 2015-01-01 |    1     |    100    |      50      |          50          |     50     |
|------------+----------+-----------+--------------+----------------------+------------|
| 2015-01-01 |    2     |    10     |      3       |          7           |     7      |
|------------+----------+-----------+--------------+----------------------+------------|
| 2015-01-01 |    3     |     0     |      0       |          0           |     0      |
|------------+----------+-----------+--------------+----------------------+------------|
| 2015-01-02 |    1     |     0     |      50      |         -50          |     0      |
|------------+----------+-----------+--------------+----------------------+------------|
| 2015-01-02 |    2     |    20     |      0       |          20          |     27     |
|------------+----------+-----------+--------------+----------------------+------------|
| 2015-01-02 |    3     |     0     |      0       |          0           |     0      |
|------------+----------+-----------+--------------+----------------------+------------|

如果您可以使用Window功能（效率更高），您可以执行以下操作：

SELECT
    month, 
    marketid, 
    totalsold, 
    totalshipped, 
    lefttoship_thismonth, 
    ( sum(lefttoship_thismonth) 
           OVER (PARTITION BY marketid ORDER BY month ROWS UNBOUNDED PRECEDING)
    ) AS total_left
FROM 
    t t1
ORDER BY
    month, marketid ;

如果您的month列是varchar（不是一个好主意），您可以将其转换为日期，或使用to_date函数。

Answer 3

评论太长了。

如果列是varchar类型，那么你必须将它转换为日期才能在order by子句中使用它。就像这样。

select t.*,
  sum(totallefttoship) over 
   (partition by marketid order by to_date(month,'dd-mm-yyyy')) as TOTALLEFT
From yourtable t

如何在PostgreSQL 9.3中对条件求和两列

3 个答案: