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SELECT语句

SELECT DISTINCT owners.name, owners.pet, owners.city,
group_concat(DISTINCT types.type separator ', ') AS type
FROM tbl_pet_owners owners
INNER JOIN tbl_pet_types types ON owners.pet = types.pet
WHERE owners.city = 'Boston' OR owners.city = 'NYC'
GROUP BY owners.name, owners.pet
ORDER BY owners.city

..返回此结果：

+--------+----------+--------+-------------------+
| name   | pet      | city   | type              |
+--------+----------+--------+-------------------+
| jack   | cat      | Boston | mammal            |
| jane   | cat      | Boston | mammal            |
| jane   | dog      | Boston | mammal            |
| jim    | goldfish | Boston | fish, seacreature |
| jim    | snake    | Boston | reptile           |
| drew   | dog      | NYC    | mammal            |
| joseph | cat      | NYC    | mammal            |
| sam    | cat      | NYC    | mammal            |
+--------+----------+--------+-------------------+

不幸的是，结果中省略了杰克的青蛙，因为在tbl_pet_types中没有青蛙的条目。如何编辑我的查询以在结果中包含jack的青蛙（使用type = NULL）？

Answer 1

您正在使用INNER JOIN，因为两张桌子都没有匹配，因此该行不会显示。尝试使用LEFT JOIN，以便左（第一）表中的所有值都会显示，无论右（第二）表上是否有答案。

您可以使用此图像对此进行可视化

Answer 2

我可以建议以下查询，这与您最初的查询相似，但稍作修改。

SELECT
    owners.name,
    owners.pet,
    owners.city,
    GROUP_CONCAT(DISTINCT COALESCE(types.type, 'NA') separator ', ') AS type
FROM tbl_pet_owners owners
LEFT JOIN tbl_pet_types types
    ON owners.pet = types.pet
WHERE owners.city = 'Boston' OR owners.city = 'NYC'
GROUP BY owners.name, owners.pet, owners.city
ORDER BY owners.city

您不需要在select子句中使用DISTINCT，因为GROUP BY已经实现了您的想法。最重要的变化是切换到LEFT JOIN。这样可以防止记录丢失。

Answer 3

对tbl_pet_types使用LEFT JOIN而不是INNER JOIN。 INNER JOIN的作用是确保您只看到两个匹配的表上的记录。在这种特殊情况下，您需要所有tbl_pet_owners数据，无论它是否与tbl_pet匹配。

SELECT DISTINCT owners.name, owners.pet, owners.city,
group_concat(DISTINCT types.type separator ', ') AS type
FROM tbl_pet_owners owners
LEFT JOIN tbl_pet_types types ON owners.pet = types.pet
WHERE owners.city = 'Boston' OR owners.city = 'NYC'
GROUP BY owners.name, owners.pet
ORDER BY owners.city

在MySQL SELECT结果中包含NULL值

3 个答案: