Question

我有两个查询，结果应该是互斥的，但事实并非如此。

这个正确查找所有具有匹配名称和order_id的电子邮件的帐户：

SELECT * FROM `accounts` 
LEFT OUTER JOIN `emails` ON emails.account_id = accounts.id 
WHERE (emails.type_name = 'name' AND emails.order_id = 1)

这个应该找到所有没有匹配名称和order_id的电子邮件的帐户，或者根本没有电子邮件：

SELECT * FROM `accounts` 
LEFT OUTER JOIN `emails` ON emails.account_id = accounts.id 
WHERE (!(emails.type_name = 'name' AND emails.order_id = 1) OR emails.id IS NULL)

但是，如果后一个查询的电子邮件不匹配，则会返回具有匹配电子邮件的帐户，因此它会从第一个查询返回帐户。任何帮助将不胜感激。

Answer 1

考虑以下测试用例：

CREATE TABLE accounts (id int);
CREATE TABLE emails (id int, account_id int, type_name varchar(10), order_id int);

INSERT INTO accounts VALUES (1), (2), (3), (4);

INSERT INTO emails VALUES (1, 1, 'name', 1);
INSERT INTO emails VALUES (2, 1, 'no-name', 1);
INSERT INTO emails VALUES (3, 2, 'name', 1);
INSERT INTO emails VALUES (4, 2, 'no-name', 1);
INSERT INTO emails VALUES (5, 3, 'name', 2);

然后按预期工作：

SELECT * FROM `accounts` 
LEFT OUTER JOIN `emails` ON emails.account_id = accounts.id 
WHERE (emails.type_name = 'name' AND emails.order_id = 1);

+------+------+------------+-----------+----------+
| id   | id   | account_id | type_name | order_id |
+------+------+------------+-----------+----------+
|    1 |    1 |          1 | name      |        1 |
|    2 |    3 |          2 | name      |        1 |
+------+------+------------+-----------+----------+
2 rows in set (0.00 sec)

第二个查询的问题是，如果帐户中没有电子邮件，则可以返回NULL行，如帐号4所示：

SELECT * FROM `accounts` 
LEFT OUTER JOIN `emails` ON emails.account_id = accounts.id 
WHERE (!(emails.type_name = 'name' AND emails.order_id = 1) OR emails.id IS NULL);

+------+------+------------+-----------+----------+
| id   | id   | account_id | type_name | order_id |
+------+------+------------+-----------+----------+
|    1 |    2 |          1 | no-name   |        1 |
|    2 |    4 |          2 | no-name   |        1 |
|    3 |    5 |          3 | name      |        2 |
|    4 | NULL |       NULL | NULL      |     NULL |
+------+------+------------+-----------+----------+
4 rows in set (0.01 sec)

为什么这对于没有NULL行的互斥结果集来说不够？：

SELECT * FROM `accounts` 
LEFT OUTER JOIN `emails` ON emails.account_id = accounts.id 
WHERE NOT (emails.type_name = 'name' AND emails.order_id = 1)

+------+------+------------+-----------+----------+
| id   | id   | account_id | type_name | order_id |
+------+------+------------+-----------+----------+
|    1 |    2 |          1 | no-name   |        1 |
|    2 |    4 |          2 | no-name   |        1 |
|    3 |    5 |          3 | name      |        2 |
+------+------+------------+-----------+----------+
3 rows in set (0.00 sec)

Answer 2

如果您的目标要匹配一对（type_name，order_id），那么它应该有效 -

SELECT * FROM `accounts` 
LEFT OUTER JOIN `emails` ON emails.account_id = accounts.id 
WHERE (emails.type_name != 'name' OR emails.order_id != 1)

检查连接表中的不匹配

2 个答案: