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有两个表：“saldos_one”和“saldos_two”，两个表都有以下字段：CLIENTID、DATE、VALUE。

我必须在某些时间段内从每个 CLIENTID 的算术平均值中获得总算术平均值。

让我们举一些例子：

> SELECT * FROM saldos_one;
+----------+------------+---------+
| CLIENTID | DATE       | VALUE   |
+----------+------------+---------+
| 1        | 2009-08-01 | 1000.00 |
| 1        | 2009-09-01 | 2000.00 |
| 1        | 2009-10-01 | 3000.00 |
| 2        | 2009-08-01 | 1000.00 |
| 2        | 2009-09-01 | 2000.00 |
| 2        | 2009-10-01 | 3000.00 |
| 3        | 2009-08-01 | 1000.00 |
| 3        | 2009-09-01 | 2000.00 |
| 3        | 2009-10-01 | 3000.00 |
| 4        | 2009-08-01 | 1000.00 |
| 4        | 2009-09-01 | 2000.00 |
| 4        | 2009-10-01 | 3000.00 |
+----------+------------+---------+

> SELECT * FROM saldos_two;
+----------+------------+---------+
| CLIENTID | DATE       | VALUE   |
+----------+------------+---------+
| 1        | 2009-08-01 | 10.00   |
| 1        | 2009-09-01 | 20.00   |
| 1        | 2009-10-01 | 30.00   |
| 2        | 2009-08-01 | 100.00  |
| 2        | 2009-09-01 | 200.00  |
| 2        | 2009-10-01 | 300.00  |
| 3        | 2009-08-01 | 1000.00 |
| 3        | 2009-09-01 | 2000.00 |
| 3        | 2009-10-01 | 3000.00 |
| 5        | 2009-08-01 | 1.00    |
| 5        | 2009-09-01 | 2.00    |
| 5        | 2009-10-01 | 3.00    |
+----------+------------+---------+

每张表的 QUERY 算术均值后：

> SELECT CLIENTID, TRUNCATE(SUM(VALUE)/COUNT(VALUE), 2) 
FROM saldos_one 
WHERE (DATE BETWEEN '2009-08-01' AND '2009-10-01')
GROUP BY CLIENTID;
+----------+---------+
| CLIENTID | VALUE   |
+----------+---------+
| 1        | 2000.00 |
| 2        | 2000.00 |
| 3        | 2000.00 |
| 4        | 2000.00 |
+----------+---------+

> SELECT CLIENTID, TRUNCATE(SUM(VALUE)/COUNT(VALUE), 2) 
FROM saldos_two
WHERE (DATE BETWEEN '2009-08-01' AND '2009-10-01')
GROUP BY CLIENTID;
+----------+---------+
| CLIENTID | VALUE   |
+----------+---------+
| 1        | 20.00   |
| 2        | 200.00  |
| 3        | 2000.00 |
| 5        | 2.00    |
+----------+---------+

我想从不同表的算术平均值中得到每个客户的算术平均值，即：

+----------+---------+
| CLIENTID | VALUE   |
+----------+---------+
| 1        | 1010.00 | = 2000.00 + 20.00 / 2
| 2        | 200.00  | = 200.00 + 200.00 / 2
| 3        | 2000.00 | = 2000.00 + 2000.00 / 2
| 4        | 1000.00 | = 2000.00 + 0 / 2
| 5        | 1.00    | = 2.00 + 0 / 2
+----------+---------+

解决方案：查看@bvr 的回复

4

4 回答 4

3

试试这个

SELECT CLIENTID,SUM(VALUE)/2 VALUE FROM
(
    SELECT CLIENTID, TRUNCATE(SUM(VALUE)/COUNT(VALUE), 2) VALUE 
    FROM saldos_one 
    WHERE (DATE BETWEEN '2009-08-01' AND '2009-10-01')
    GROUP BY CLIENTID;
    UNION ALL
    SELECT CLIENTID, TRUNCATE(SUM(VALUE)/COUNT(VALUE), 2) VALUE
    FROM saldos_two
    WHERE (DATE BETWEEN '2009-08-01' AND '2009-10-01')
) t
 GROUP BY CLIENTID
于 2013-08-27T10:04:45.470 回答
1

您可以使用其中UNION的一个表，然后对它们执行分析。

例如

SELECT CLIENTID, AVG(VALUE) FROM (
    SELECT CLIENTID, VALUE FROM saldos_one WHERE 
       (DATE BETWEEN '2009-08-01' AND '2009-10-01')
  UNION
    SELECT CLIENTID, VALUE FROM saldos_two WHERE 
       (DATE BETWEEN '2009-08-01' AND '2009-10-01')
) t
GROUP BY CLIENTID
于 2013-08-27T10:05:42.517 回答
1 
select CLIENTID, TRUNCATE(SUM(VALUE)/COUNT(VALUE), 2)  from
(SELECT CLIENTID, TRUNCATE(SUM(VALUE)/COUNT(VALUE), 2)  as VALUE
FROM saldos_one 
WHERE (DATE BETWEEN '2009-08-01' AND '2009-10-01')
GROUP BY CLIENTID
union all
 SELECT CLIENTID, TRUNCATE(SUM(VALUE)/COUNT(VALUE), 2) 
FROM saldos_two
WHERE (DATE BETWEEN '2009-08-01' AND '2009-10-01')
GROUP BY CLIENTID) as A group by A.CLIENTID
于 2013-08-27T10:09:29.057 回答
0

您可以使用 AVG 函数代替 SUM(VALUE)/COUNT(VALUE)。

关于你问题的最后一部分：

您可以将两个查询的结果放入临时表中，然后使用 AVG 函数将同一查询应用于临时表，因此您可以在两个值（2000 和 20）之间应用平均值。

和你桌子的所有行

于 2013-08-27T10:02:40.127 回答