我的查询

Question

数据库的表结构：
教练表

+----+-----------+-------+  
| ID |   Name    | Age   |  
+----+-----------+-------+  
|  1 | Aaron     |  39   |  
|  2 | John      |  41   |  
|  3 | Macy      |  44   |  
|  4 | Mitchelle |  37   |  
|  5 | Candice   |  32   |  
+----+-----------+-------+

实习生表

+----+---------+-------+  
| ID |  Name   | Age   |  
+----+---------+-------+  
|  6 | Abigail |   9   |  
|  7 | Jim     |  12   |  
|  8 | Jack    |   7   |  
|  9 | Maria   |  14   |  
| 10 | Andy    |   11  |  
+----+---------+-------+

费用表

+----+----------+------------+--------+  
| ID | Coach_ID | Trainee_ID | Fee    |  
+----+----------+------------+--------+  
| 11 |        1 |          7 | 2400   |  
| 12 |        2 |          6 | 2000   |  
| 13 |        3 |          6 | 2000   |  
| 14 |        4 |          8 | 1243   |  
| 15 |        5 |          8 | 1275   |  
| 16 |        3 |          9 | 9010   |  
| 17 |        2 |          8 | 1900   |  
| 18 |        1 |          7 |  600   |  
| 19 |        2 |         10 | 1010   |  
| 20 |        5 |         10 |   2110 |  
+----+----------+------------+--------+

必需的输出

+-----------+--------------+--------+  
|   Name    | Trainee_name | Fee    |  
+-----------+--------------+--------+  
| Aaron     | Jim          | 2400   |  
| Candice   | Andy         | 2110   |  
| John      | Abigail      | 2000   |  
| Macy      | Maria        | 9010   |  
| Mitchelle | Jack         |  1243  |  
+-----------+--------------+--------+

=====

我的查询

Select 
 Coach.Name as Name, Trainee.Name as Trainee_name, Max(Fee.Fee) as Fee 
From Fee
 INNER JOIN Trainee
  ON Fee.Trainee_ID = Trainee.ID
 INNER JOIN Coach
  ON Fee.Coach_ID = Coach.ID 
GROUP BY Coach.Name
ORDER BY Coach.Name;

我的输出：

+------------+---------------+--------+  
|   Name     | Trainee_name  | Fee    |  
+------------+---------------+--------+  
| Aaron      | Jim           | 2400   |  
| Candice    | Jack          | 2110   |  
| John       | Abigail       | 2000   |  
| Macy       | Abigail       | 9010   |  
| Mitchelle  | Jack          |  1243  |  
+------------+---------------+--------+

对于Candice和Macy，对应的数据在trainee_name列中不正确。

转储sql数据库：

CREATE TABLE IF NOT EXISTS `coach` (  
  `ID` int(11) NOT NULL,  
  `Name` text NOT NULL,  
  `Age` int(11) NOT NULL  
) ENGINE=InnoDB DEFAULT CHARSET=latin1;  


INSERT INTO `coach` (`ID`, `Name`, `Age`) VALUES  
(1, 'Aaron', 39),  
(2, 'John', 41),  
(3, 'Macy', 44),  
(4, 'Mitchelle', 37),  
(5, 'Candice', 32);  


CREATE TABLE IF NOT EXISTS `fee` (  
  `ID` int(11) NOT NULL,  
  `Coach_ID` int(11) NOT NULL,  
  `Trainee_ID` int(11) NOT NULL,  
  `Fee` int(11) NOT NULL  
) ENGINE=InnoDB DEFAULT CHARSET=latin1;  


INSERT INTO `fee` (`ID`, `Coach_ID`, `Trainee_ID`, `Fee`) VALUES  
(11, 1, 7, 2400),  
(12, 2, 6, 2000),  
(13, 3, 6, 2000),  
(14, 4, 8, 1243),  
(15, 5, 8, 1275),  
(16, 3, 9, 9010),  
(17, 2, 8, 1900),  
(18, 1, 7, 600),  
(19, 2, 10, 1010),  
(20, 5, 10, 2110);  

CREATE TABLE IF NOT EXISTS `trainee` (  
  `ID` int(11) NOT NULL,  
  `Name` text NOT NULL,  
  `Age` int(11) NOT NULL  
) ENGINE=InnoDB DEFAULT CHARSET=latin1;  


INSERT INTO `trainee` (`ID`, `Name`, `Age`) VALUES  
(6, 'Abigail', 9),  
(7, 'Jim', 12),  
(8, 'Jack', 7),  
(9, 'Maria', 14),  
(10, 'Andy', 11);

Answer 1

试试这个：

SELECT sub.name, trainee.name, sub.mx
FROM
(SELECT Fee.Coach_ID, Coach.Name, MAX(fee) AS mx
 FROM Coach
 INNER JOIN Fee ON Coach.ID = Fee.Coach_ID
 GROUP BY Fee.Coach_ID, Coach.Name) sub
INNER JOIN fee ON sub.coach_ID = fee.coach_ID
INNER JOIN trainee ON fee.trainee_ID = trainee.id
WHERE sub.mx = fee.fee
ORDER BY sub.name

您可以看到我在此处运行的测试：http://sqlfiddle.com/#!9/cdbad/10。

它有你想要的输出。诀窍是子查询。您需要首先确定每个教练的最高费用。然后通过费用表加入学员。

加入返回不正确的值

我的查询

1 个答案: