我的 foos 表中有以下数据结构:
-----------------------------------------------
| id | bar_id | baz_id | date | value |
-----------------------------------------------
| 1 | 1 | 1 | 2013-12-01 | failure |
| 2 | 1 | 1 | 2013-12-09 | failure |
| 3 | 2 | 1 | 2013-12-02 | success |
| 4 | 3 | 1 | 2013-12-10 | success |
| 5 | 3 | 1 | 2013-12-01 | failure |
| 6 | 3 | 1 | 2013-12-08 | success |
| 7 | 1 | 2 | 2013-12-02 | success |
| 8 | 1 | 2 | 2013-12-08 | failure |
| 9 | 1 | 2 | 2013-12-03 | success |
| 10 | 2 | 2 | 2013-12-07 | failure |
| 11 | 2 | 2 | 2013-12-08 | failure |
| 12 | 3 | 2 | 2013-12-04 | success |
| 13 | 3 | 3 | 2013-12-14 | failure |
-----------------------------------------------
我的目标是为不同baz_ids的每个bar_id获得成功/总数。例如:
------------------------------
| bar_id | successes | total |
------------------------------
| 1 | 1 | 2 |
| 2 | 1 | 2 |
| 3 | 2 | 3 |
------------------------------
下面是一个有效的查询:
SELECT foos.bar_id,
successes,
COUNT(distinct baz_id) as total
FROM foos
LEFT JOIN
(SELECT bar_id, count(distinct baz_id) as successes
FROM foos
WHERE value = "success"
GROUP BY bar_id) as other
ON foos.bar_id = other.bar_id
GROUP BY bar_id
有没有办法使用 MySQL 函数获取成功列而无需进行子选择?似乎必须有一种方法可以使用GROUP_CONCAT
或其他按功能分组之一来执行此操作。
编辑
使用 SUM(value="success")
很接近,但会计算不同baz_id的所有成功,而不是只计算单个成功:
SELECT bar_id,
SUM(value="success") AS successes,
COUNT(distinct baz_id) as total
FROM foos
GROUP BY bar_id
------------------------------
| bar_id | successes | total |
------------------------------
| 1 | 2 | 2 | <- Successes should be 1
| 2 | 1 | 2 |
| 3 | 3 | 3 | <- Successes should be 2
------------------------------
您可以使用带有DISTINCT
的CASE
来获得相同的结果;
SELECT bar_id,
COUNT(DISTINCT CASE WHEN value='success' THEN baz_id ELSE NULL END) successes,
COUNT(DISTINCT baz_id) total
FROM foos
GROUP BY bar_id;
用于测试的 SQL小提琴。