Question

我有一张桌子：

------------------------
|id|p_id|desired|earned|
------------------------
|1 | 1  |  5    |  7   |
|2 | 1  |  15   |  0   |
|3 | 1  |  10   |  0   |
|4 | 2  |  2    |  3   |
|5 | 2  |  2    |  3   |
|6 | 2  |  2    |  3   |
------------------------

我需要进行一些计算，并尝试在一个非常复杂的请求中进行计算，否则我知道如何使用请求数来计算它。我需要结果表如下：

---------------------------------------------------------
|p_id|total_earned|    AVG   |      Count     |  SUM    |
|    |            | (desired)|(if earned != 0)|(desired)|
---------------------------------------------------------
|  1 |      7     |     10   |       1        |    30   |
|  2 |      9     |      2   |       3        |    6    |
---------------------------------------------------------

我到目前为止构建了：

SELECT p_id, SUM(earned), AVG(desired), Sum(desired) 
FROM table GROUP BY p_id

但我无法弄清楚如何计算有条件的分组记录数。我可以使用HAVING获取此号码，但是在单独的请求中。

我几乎可以肯定SQL应该拥有这种能力。

Answer 1

您可以使用CASE表达式。

试试这个，

SELECT p_id
    ,SUM(earned) AS total_earned
    ,AVG(desired) AS avg_desired
    ,COUNT(CASE WHEN Earned!=0 THEN 1 END) AS earned_count
    ,SUM(desired) AS sum_desired
FROM table
GROUP BY p_id;

Answer 2

您几乎已经完成了查询，只需在case表达式的帮助下添加条件聚合以获得收入计数

SELECT 
           p_id,
           SUM(earned) [total_earned],
           AVG(desired) [desired],
           SUM(CASE WHEN earned <> 0 THEN 1 ELSE 0 END) [COUNT],
           SUM(desired) [SUM] FROM <table>
 GROUP BY p_id

结果

p_id    total_earned  desired   COUNT  SUM
1       7             10        1      30
2       9             2         3      6

Answer 3

CASE的缩写替代

SELECT p_id,
    SUM(earned) AS total_earned,
    AVG(desired) AS average_desired,
    COUNT(earned != 0 OR NULL) AS earned_count,
    SUM(desired) AS sum_desired
FROM table GROUP BY p_id;

因为NULL不计算在内。

Answer 4

NULLIF()是标准SQL，可能最短：

SELECT p_id
     , count(NULLIF(earned, 0)) AS earned_count
  -- , more ...
FROM   table
GROUP  BY 1;

count()仅计算非空值。

更多变种：

For absolute performance, is SUM faster or COUNT?

是否有可能将Group by，Having和Sum结合起来？

4 个答案: