Left Join/IS NULL如何消除一个表中存在而另一表中没有的记录? [英] How does Left Join / IS NULL eliminate records which are there in one table and not in the other?

问题描述

我很难理解LEFT JOIN/IS NULL为什么要消除在一个表中而不是另一个表中的记录. 这是一个例子

I am having a tough time to understand why does LEFT JOIN / IS NULL eliminate records which are there in one table and not in the other. Here is an example

SELECT  l.id, l.value
FROM    t_left l
LEFT JOIN t_right r
ON      r.value = l.value
WHERE   r.value IS NULL

为什么r.value = NULL应该消除记录?我不理解 .我知道我缺少一些非常基本的东西，但目前我还无法弄清楚那个基本的东西.如果有人向我详细解释，我将不胜感激.

Why should r.value = NULL eliminate records ? I am not understanding . I know I am missing something very basic but at present I cant figure out even that basic one. I would appreciate if someone explains it to me in detail .

我想要一个非常基本的解释.

I want a very basic explanation.

推荐答案

可以通过以下内容对此进行解释

This could be explained with the following

mysql> select * from table1 ;
+------+------+
| id   | val  |
+------+------+
|    1 |   10 |
|    2 |   30 |
|    3 |   40 |
+------+------+
3 rows in set (0.00 sec)

mysql> select * from table2 ;
+------+------+
| id   | t1id |
+------+------+
|    1 |    1 |
|    2 |    2 |
+------+------+
2 rows in set (0.00 sec)

此处table1.id <-> table2.t1id

现在，当我们使用连接键执行left join时，如果左边的表是table1，那么它将从table1获取所有数据，并且在table2的不匹配记录中将被设置为null

Now when we do a left join with the joining key and if the left table is table1 then it will get all the data from table1 and in non-matching record on table2 will be set to null

mysql> select t1.* , t2.t1id from table1 t1 
left join table2 t2 on t2.t1id = t1.id ;
+------+------+------+
| id   | val  | t1id |
+------+------+------+
|    1 |   10 |    1 |
|    2 |   30 |    2 |
|    3 |   40 | NULL |
+------+------+------+

3 rows in set (0.00 sec)

看到table1.id = 3在table2中没有值，因此将其设置为null 当您应用where条件时，它将做进一步的过滤

See that table1.id = 3 does not have a value in table2 so its set as null When you apply the where condition it will do further filtering

mysql> select t1.* , t2.t1id from table1 t1 
left join table2 t2 on t2.t1id = t1.id where t2.t1id is null;
+------+------+------+
| id   | val  | t1id |
+------+------+------+
|    3 |   40 | NULL |
+------+------+------+
1 row in set (0.00 sec)

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