Question

我有以下MySQL表：

tbl_pet_owners：

+----+--------+----------+--------+--------------+
| id | name   | pet      | city   | date_adopted |
+----+--------+----------+--------+--------------+
|  1 | jane   | cat      | Boston | 2017-07-11   |
|  2 | jane   | dog      | Boston | 2017-07-11   |
|  3 | jane   | cat      | Boston | 2017-06-11   |
|  4 | jack   | cat      | Boston | 2016-07-11   |
|  5 | jim    | snake    | Boston | 2017-07-11   |
|  6 | jim    | goldfish | Boston | 2017-07-11   |
|  7 | joseph | cat      | NYC    | 2016-07-11   |
|  8 | sam    | cat      | NYC    | 2017-07-11   |
|  9 | drew   | dog      | NYC    | 2016-07-11   |
| 10 | jack   | frog     | Boston | 2017-07-19   |
+----+--------+----------+--------+--------------+

tbl_pet_types：

+----------+-------------+
| pet      | type        |
+----------+-------------+
| cat      | mammal      |
| dog      | mammal      |
| goldfish | fish        |
| goldfish | seacreature |
| snake    | reptile     |
+----------+-------------+

我有以下SQL查询，它会搜索宠物类型列表，以及拥有这些宠物类型的所有者的名称和城市：

SELECT DISTINCT types.type, owners.name, owners.city
FROM tbl_pet_owners owners
LEFT JOIN tbl_pet_types types ON owners.pet = types.pet
WHERE types.type IN ('mammal', 'fish', 'amphibian', 'seacreature');

查询返回：

+-------------+--------+--------+
| type        | name   | city   |
+-------------+--------+--------+
| mammal      | jane   | Boston |
| mammal      | jack   | Boston |
| fish        | jim    | Boston |
| seacreature | jim    | Boston |
| mammal      | joseph | NYC    |
| mammal      | sam    | NYC    |
| mammal      | drew   | NYC    |
+-------------+--------+--------+

结果中省略了两栖动物，因为在tbl_pet_types中找不到它。如何更改我的查询以便行：

amphibian, NULL, NULL

包含在结果中？

Answer 1

left join是正确的，但您需要types成为第一个：

SELECT DISTINCT types.type, owners.name, owners.city
FROM tbl_pet_types types LEFT JOIN
     tbl_pet_owners owners
     ON owners.pet = types.pet
WHERE types.type IN ('mammal', 'fish', 'amphibian', 'seacreature');

由于WHERE子句仅指tbl_pet_types，因此不会更改。

left join的工作原理很简单：它会将所有行保留在第一个表中。第二个中不匹配的列变为NULL。

编辑：

如果您有一个不在tbl_pet_types中的类型列表，那么您需要left join包含派生表中的所有值：

SELECT DISTINCT tt.type, po.name, po.city
FROM (SELECT 'mammal' as type UNION ALL
      SELECT 'fish' as type UNION ALL
      SELECT 'amphibian' as type UNION ALL
      SELECT 'seacreature' as type 
     ) tt left join
     tbl_pet_types pt
     ON pt.type = tt.type LEFT JOIN
     tbl_pet_owners po
     ON po.pet = pt.pet;

Answer 2

然后执行RIGHT JOIN而不是LEFT JOIN说

RIGHT JOIN tbl_pet_types types ON owners.pet = types.pet

如果你想要两个表的外连接结果（不确定关闭），你可以考虑通过混合左右连接来做FULL JOIN

Answer 3

MySQL STILL 没有CTE吗？它们让生活变得更加轻松。

无论如何，要获得“未找到”，请在SELECT中使用SELECT DISTINCT p.petType , coalesce(po.name,'NOT FOUND') AS name , coalesce(po.city,'NOT FOUND') AS city FROM ( SELECT 'mammal' AS petType UNION ALL SELECT 'fish' UNION ALL SELECT 'amphibian' UNION ALL SELECT 'seacreature' ) p LEFT OUTER JOIN ( SELECT tpo.name, tpo.city, pt.type AS petType FROM tbl_pet_owners tpo LEFT OUTER JOIN tbl_pet_types tpt ON tpo.pet = tpt.pet ) po ON p.petType = po.petType。

{{1}}

我没有MySQL机器来测试这个，但像这样的东西应该为你提供你想要的东西。

在MySQL查询结果中包含NOT FOUND值

3 个答案: