Question

我创建了以下简单表格来说明我们要解决的问题：

  CREATE TABLE soPost
  (
    id    INT AUTO_INCREMENT,
    code  INT,
    value VARCHAR(8),

    PRIMARY KEY(id)
  );

该表已填充以下插入语句：

INSERT INTO soPost(code, value)
VALUES (101, 'XYZ'), (101, 'ABC'), (101, 'DEF'), (101, 'GHI'),
       (201, 'XYZ'), (201, 'ABC'), (201, 'JKL'),
       (301, 'XYZ'), (301, 'DEF'), (301, 'MNO'),
       (401, 'XYZ'), (401, 'PRQ'),
       (501, 'STU'), (501, 'VWZ');

结果

SELECT * FROM soPost;

是：

+----+------+-------+
| id | code | value |
+----+------+-------+
|  1 |  101 | XYZ   |
|  2 |  101 | ABC   |
|  3 |  101 | DEF   |
|  4 |  101 | GHI   |
|  5 |  201 | XYZ   |
|  6 |  201 | ABC   |
|  7 |  201 | JKL   |
|  8 |  301 | XYZ   |
|  9 |  301 | DEF   |
| 10 |  301 | MNO   |
| 11 |  401 | XYZ   |
| 12 |  401 | PRQ   |
| 13 |  501 | STU   |
| 14 |  501 | VWZ   |
+----+------+-------+

我们需要选择其中至少一个值是“ XYZ”但都不是“ ABC”或“ DEF”的代码。显然，满足此条件的唯一值是代码401。

请提出SQL查询以实现此结果。

Answer 1

您可以使用NOT EXISTS子查询来执行此测试：

SELECT code 
FROM soPost s1
WHERE s1.value = 'XYZ' AND
  NOT EXISTS (SELECT * 
              FROM soPost s2 
              WHERE s2.code = s1.code AND s2.value IN('ABC', 'DEF'))

输出：

code
401

Answer 2

您可以使用子数据集： select distinct(code) from soPost where code not in ( select distinct(code) from soPost where value = "ABC" or value = "DEF" ) and code in ( select code from soPost where value='XYZ' );

MySQL ---满足多个元组条件的SELECT查询

2 个答案: