Question

这吓坏了我！得到以下数据：

+----+-----+-------+------------+
| ID | REG | VALUE | DATE       |
+----+-----+-------+------------+
| 1  | 1A  | 100   | 2009-01-01 |
| 1  | 1A  | 100   | 2009-02-01 |
| 1  | 1A  | 100   | 2009-03-01 |
| 2  | 1B  | 100   | 2009-01-01 |
| 2  | 1B  | 100   | 2009-02-01 |
| 2  | 1B  | 100   | 2009-03-01 |
| 2  | 1C  | 100   | 2009-01-01 |
| 2  | 1C  | 100   | 2009-02-01 |
| 2  | 1C  | 200   | 2009-03-01 |
+----+-----+-------+------------+

PS {edit 0001} ::还有一个额外的字段，也必须用于过滤数据，称之为{TYPE}，可以将'SINGLE'或'MULTIPLE'作为值。

我想在每个{ID}的SUM（每个不同的{REG}）之间获得MAX。显然，这是一个简单的表示，表达到了64985928个寄存器，{DATE}是过滤数据。

那将是第一步获得每个{REG}的SUM：

+----+------+
| ID | SUM  |
+----+------+
| 1  | 300  |
| 2  | 300  |
| 2  | 400  |
+----+------+

那是：

SELECT 
  SUM(value) 
FROM 
  table 
WHERE
  (date BETWEEN '2009-01-01' AND '2009-03-01')
GROUP BY
  reg;

然后，从每个SUM中获取MAX，这是我被困的地方：

+----+------+
| ID | MAX  |
+----+------+
| 1  | 300  |
| 2  | 400  |
+----+------+

我试过了：

SELECT
  a.id,
  MAX(b.sum)
FROM
  table a,
  (SELECT 
     SUM(b.value) 
   FROM 
     table b 
   WHERE 
     (b.date BETWEEN '2009-01-01' AND '2009-03-01') AND (a.id = b.id)
   GROUP BY
     b.reg);

有什么想法吗？ PS：对不起错误。

PS {edit 0002} 要复制原始查询和数据，这样可能会有所帮助。

$ QUERY：

SELECT 
  clienteid AS "CLIENTE",
  SUM(saldo) AS "SUMA" 
FROM
  etl.creditos
WHERE
   (titularidad_tipo LIKE 'TITULAR')
AND
   (mes_datos BETWEEN '2008-11-01' AND '2009-10-01')
GROUP BY
  nuc 
ORDER BY
  clienteid;

GOT：

+---------+-------------+
| CLIENTE | SUMA        |
+---------+-------------+
| 64      | 1380690.74  |
| 187     | 1828468.71  |
| 187     | 2828102.80  |
| 325     | 26037422.21 |
| 389     | 875519.05   |
| 495     | 20084.93    |
| 495     | 109850.46   |
+---------+-------------+

然后，我正在寻找的是：

+---------+-------------+
| CLIENTE | MAX         |
+---------+-------------+
| 64      | 1380690.74  |
| 187     | 1828468.71  |
| 325     | 26037422.21 |
| 389     | 875519.05   |
| 495     | 109850.46   |
+---------+-------------+

但是跑步：

SELECT
    clienteid AS "CLIENTE",
    MAX(suma)
FROM
    (SELECT clienteid, SUM(saldo) AS "suma" FROM etl.creditos
    WHERE (mes_datos BETWEEN '2009-08-01' AND '2009-10-01') AND (titularidad_tipo LIKE 'TITULAR')
    GROUP BY clienteid, nuc) AS sums
GROUP BY
    clienteid
ORDER BY
    clienteid;

结果为：

+---------+-------------+
| CLIENTE | SUMA        |
+---------+-------------+
| 64      | 336879.21   |
| 187     | 1232824.51  |
| 325     | 3816173.62  |
| 389     | 218423.83   |
| 495     | 34105.99    |
+---------+-------------+

Answer 1

SELECT ID, MAX(reg_sum)
FROM
(
   SELECT ID, SUM(value) AS reg_sum FROM table 
   WHERE  (date BETWEEN '2009-01-01' AND '2009-03-01')
   GROUP BY  ID, reg
) a GROUP by ID

Answer 2

您可以添加[按SUM（值）DESC限制1的顺序]以获取查询结果的最大值。

 SELECT  SUM(value) as maxcount  FROM table WHERE (date BETWEEN '2009 01-01' AND '2009-03-01') GROUP BY reg  order by maxcount desc  limit 1;

来自SUM的MySQL MAX

2 个答案: