Question

我在这里有疑问......

Table : history
|id   |transaction|created_at         |merchant_id|
|-----|-----------|-------------------|-----------|
|1    |400        |2015-10-12 11:08:37|33         |
|1    |500        |2015-10-15 09:38:22|33         |
|1    |600        |2015-10-21 14:47:12|22         |
|2    |100        |2015-09-26 10:48:27|31         |
|2    |500        |2015-09-30 11:18:07|27         |
|2    |300        |2015-10-02 17:33:57|31         |

我希望当我做查询时：

SELECT SUM(a.transaction)/COUNT(a.transaction) AS avg_trans
FROM history AS a GROUP BY a.id, a.merchant_id

Result:
|id    |avg_trans|merchant_id|
|------|---------|-----------|
|1     |450      |33         |
|1     |600      |22         |
|2     |200      |31         |
|2     |500      |27         |

然后将avg_trans显示到表历史记录中，如下所示：

|id   |transaction|created_at         |avg_trans|merchant_id|
|-----|-----------|-------------------|---------|-----------|
|1    |400        |2015-10-12 11:08:37|450      |33         |
|1    |500        |2015-10-15 09:38:22|450      |33         |
|1    |600        |2015-10-21 14:47:12|600      |22         |
|2    |100        |2015-09-26 10:48:27|200      |31         |
|2    |200        |2015-09-30 11:18:07|500      |27         |
|2    |300        |2015-10-02 17:33:57|200      |31         |

任何人都可以帮助我吗？

Answer 1

正如@Strawberry指出的那样，您需要使用子查询来计算ID和商家ID的平均值，并将其加入主表：

select t1.*, t2.avg_trans
from table t1
inner join (select id, merchant_id, avg(transaction) avg_trans
            from table
            group by id, merchant_id) t2 on t1.id=t2.id and t1.merchant_id=t2.merchant_id

Answer 2

 select history.*, derived_table.avg_trans from history left join (SELECT id, merchant_id, SUM(a.transaction)/COUNT(a.transaction) AS avg_trans FROM history AS a GROUP BY a.id, a.merchant_id) derived_table
on history.merchant_id = derived_table.merchant_id

选择Join table with table self

2 个答案: