mysql - 选择外行仅匹配单个值的行

Question

假设我有两个表——people 和 pets——每个人可能有不止一只宠物:

人:

+-----------+-------+
| person_id | name  |
+-----------+-------+
|         1 | Bob   |
|         2 | John  |
|         3 | Pete  |
|         4 | Waldo |
+-----------+-------+

宠物:

+--------+-----------+--------+
| pet_id | person_id | animal |
+--------+-----------+--------+
|      1 |         1 | dog    |
|      2 |         1 | dog    |
|      3 |         1 | cat    |
|      4 |         2 | cat    |
|      5 |         3 | dog    |
|      6 |         3 | tiger  |
|      7 |         3 | tiger  |
|      8 |         4 | tiger  |
|      9 |         4 | tiger  |
|     10 |         4 | tiger  |
+--------+-----------+--------+

我正在尝试选择只拥有老虎作为宠物的人。显然，唯一符合此标准的是 Waldo，因为 Pete 也有一只 dog...但我在编写时遇到了一些问题对此的查询。

最明显的例子是select people.person_id, people.name from people join pets on people.person_id = pets.person_id where pets.animal = "tiger"，但这返回皮特 和沃尔多。

如果有像 pets.animal ONLY = "tiger" 这样的子句会很有帮助，但据我所知这并不存在。

如何编写查询？

Answer 1

 select people.person_id, people.name 
   from people 
   join pets on people.person_id = pets.person_id 
  where pets.animal = "tiger" 
    AND people.person_id NOT IN (select person_id from pets where animal != 'tiger');