包含空值在MySQL SELECT结果

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  | 
+----------+-------------+

我有下面的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

..这将返回这样的结果：因为在tbl_pet_types青蛙没有进入

+--------+----------+--------+-------------------+ 
| 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，杰克的青蛙从结果中省略。我如何编辑我的查询以在结果中包含jack的青蛙（type = NULL）？

来源

2017-07-19 Brinley

将你的'INNER JOIN'改为'LEFT OUTER JOIN'并让'呃撕裂。 'INNER JOIN'说“只在连接条件为真时返回结果”。 'LEFT OUTER JOIN'说：“返回第一个表中的所有结果，仅返回条件为真的左连接表中的结果” – JNevill

[INNER JOIN]与“OUTER JOIN”之间的区别是什么“？]（https://stackoverflow.com/questions/38549/what-is-the-difference-between-inner-join-and-outer-join） – JNevill

@JNevill我认为'WHERE'条款中的那些条件也是需要移到'ON'子句。 –

想象这个使用LEFT JOIN到tbl_pet_types而不是INNER JOIN。 INNER JOIN的功能是确保您只能看到两个表中匹配的记录。在这种特殊情况下，无论tbl_pet_owners数据是否与tbl_pet匹配，都需要所有tbl_pet_owners数据。

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

来源

2017-07-19 15:13:50 Isaiah3015

不需要在select子句中使用'DISTINCT'，并且'owners.city'应该包含在'GROUP BY'列表中。 –

我可以建议以下查询，它类似于你最初的，但有一些修改。

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

你不需要在你的SELECT子句中使用DISTINCT，因为GROUP BY就已经达到你脑子里想的是什么。最主要的变化是切换到LEFT JOIN。这应该防止记录被丢弃。

来源

2017-07-19 15:13:04

您使用的是INNER JOIN，因为两个表上都没有匹配，所以不会显示该行。尝试使用LEFT JOIN，以便左侧（第一个）表中的所有值都将显示出来，而不管右侧（第二个）表上是否有答案。

你可以用这个形象

来源

2017-07-19 15:13:41

包含空值在MySQL SELECT结果

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