我有以下3个表:单位,阶段,统计数据。
unit stage
+----+--------+ +----+-------+---------------------+
| id | status | | id |unit_id| date |
+----+--------+ +----+-------+---------------------+
| 1 | 2 | | 1 | 2 | 2013-11-22 00:00:00 |
| 2 | 3 | | 2 | 2 | 2013-11-26 12:00:00 |
| 3 | 3 | | 3 | 3 | 2013-10-11 00:00:00 |
| 4 | 0 | | 4 | 1 | 2013-12-29 00:00:00 |
+----+--------+ +----+-------+---------------------+
stats
+----+----------+---------------------+-------+
| id | stage_id | date | clicks|
+----+----------+---------------------+-------+
| 1 | 1 | 2013-11-22 00:00:00 | 10 |
| 2 | 1 | 2013-11-23 00:00:00 | 20 |
| 3 | 1 | 2013-11-24 00:00:00 | 25 |
| 4 | 2 | 2013-11-26 00:00:00 | 15 |
| 5 | 2 | 2013-11-27 12:00:00 | 21 |
| 6 | 3 | 2013-12-29 00:00:00 | 8 |
+----+----------+---------------------+-------+
我需要一个请求,它会产生以下响应:
+---------+---------------------+-----------------------+
| unit.id | stage.min.date | sum(stats.max.clicks) |
+---------+---------------------+-----------------------+
| 2 | 2013-11-22 00:00:00 | 46 |
| 3 | 2013-12-29 00:00:00 | 8 |
+---------+---------------------+-----------------------+
遵守以下规则:
1)unit.id
- 仅显示unit.status=3
2)stage.min.date
- 对应stage.date
unit_id
3)sum(stats.max.clicks)
- stats.clicks
的总和,其中每个stage_id
的最大点数与相应的unit_id
相关联。在我的示例中,46 = 25(stage_id=1
)+ 21(stage_id=2
)
问题在于min.date和点击总和 - 我不知道如何在一个查询中获取它。使用PHP代码和几个请求肯定不是问题。
提前致谢。
答案 0 :(得分:1)
我只是问自己,为什么我这样做?你的例子共鸣有一个错误,与你的小提琴不符......但是:
SELECT
cc.unit_id, MIN(cc.date) as stage_min_date , SUM(dd.clicks) as stats_max_clicks
FROM
stage cc
LEFT JOIN (
SELECT
bb.stage_id, bb.clicks
FROM
stats bb LEFT JOIN (
SELECT id, stage_id, MAX(date) AS max_date
FROM stats
GROUP BY stage_id
) aa
ON
aa.max_date = bb.date
WHERE
aa.max_date IS NOT NULL
) dd
ON cc.id = dd.stage_id
LEFT JOIN unit ee
ON ee.id = cc.unit_id
WHERE ee.status = 3
GROUP BY cc.unit_id
...