Question

我有3张桌子

+----+---------+
| id | message |
+----+---------+
| 1  | sup     |
+----+---------+
| 2  | hello   |
+--------------+

+------------+----------+
| message_id | comments |
+------------+----------+
| 1          | hi       |
+------------+----------+ 
| 1          | bye      |
+------------+----------+ 

+------------+-------+
| message_id | likes |
+------------+-------+
| 1          | sup   |
+------------+-------+
| 2          | hello |
+------------+-------+

我最终会找到像这样的东西

+----+---------+------------+----------+------------+-------+
| id | message | message_id | comments | message_id | likes |
+----+---------+------------+----------+------------+-------+
| 1  | sup     | 1          | hi       | null       | null  |
+----+---------+------------+----------+------------+-------+
| 1  | sup     | 1          | bye      | null       | null  |
+----+---------+------------+----------+------------+-------+
| 1  | hello   | null       | null     | 1          | sup   |
+----+---------+------------+----------+------------+-------+
| 1  | hello   | null       | null     | 2          | hello |
+----+---------+------------+----------+------------+-------+

我有2个连接，比如

select * from message join comments on id = comments.message_id join likes on id = likes.message_id

但这会返回填充的所有字段。我希望字段不会在不存在的地方填充。这可能吗？

编辑：解决人们的意见

mysql> select * from message join comments on id = comments.message_id join likes on id = likes.message_id;
+----+---------+------------+----------------------+------------+-----------+
| id | message | message_id | comment              | message_id | comment   |
+----+---------+------------+----------------------+------------+-----------+
|  1 | hello   |          1 | hello                |          1 | what      |
|  1 | hello   |          1 | hello                |          1 | what what |
|  2 | two     |          2 | bye                  |          2 | what ok   |
|  1 | hello   |          1 | hello what what      |          1 | what      |
|  1 | hello   |          1 | hello what what      |          1 | what what |
|  1 | hello   |          1 | hello what what what |          1 | what      |
|  1 | hello   |          1 | hello what what what |          1 | what what |
+----+---------+------------+----------------------+------------+-----------+

mysql> select * from message left join comments on id = comments.message_id left join likes on id = likes.message_id;
+----+---------+------------+----------------------+------------+-----------+
| id | message | message_id | comment              | message_id | comment   |
+----+---------+------------+----------------------+------------+-----------+
|  1 | hello   |          1 | hello                |          1 | what      |
|  1 | hello   |          1 | hello what what      |          1 | what      |
|  1 | hello   |          1 | hello what what what |          1 | what      |
|  1 | hello   |          1 | hello                |          1 | what what |
|  1 | hello   |          1 | hello what what      |          1 | what what |
|  1 | hello   |          1 | hello what what what |          1 | what what |
|  2 | two     |          2 | bye                  |          2 | what ok   |
+----+---------+------------+----------------------+------------+-----------+

我同意喜欢应该是一个整数，但这些数据是我刚才编写的模拟数据。

Answer 1

您只需使用LEFT JOIN代替INNER JOIN：

SELECT     *
FROM       message  M
LEFT JOIN  comments C  ON M.id = C.message_id
LEFT JOIN  likes    L  ON M.id = L.message_id

有关不同类型连接的更多信息，请访问：

What's the difference between INNER JOIN, LEFT JOIN, RIGHT JOIN and FULL JOIN?

Answer 2

也许尝试一下，看看它是否有帮助;

select * from message 
left join comments on id = comments.message_id 
left join likes on id = likes.message_id

如需进一步阅读，请查看; http://www.w3schools.com/Sql/sql_join_left.asp

Answer 3

是的，左连接会将所有行保留在该连接表达式的左表或第一个表中，所以这可能是你想要的：

SELECT * 来自消息 LEFT JOIN注释ON id = comments.message_id LEFT JOIN喜欢ON id = likes.message_id

如何在没有数据的情况下进行两次连接并显示空值

3 个答案: