Mysql在In子句中缺少值

Question

我想在查询的in子句中缺少值

人

ID    |  Name | location_id | 
1     |    Rex|          5  |
2     |    Tex|          7  |
3     |    Jim|          8  |
4     |   John|          11 |
5     |    Tom|          12 |

放置

location_id | Place    |
5           | New York |
6           | London   |
7           | Madrid   |
8           | Paris    |   
9           | Dublin   |   
10          | Chicago  |   
11          | Berlin   |   
12          | Colombo  |   

我想查找没有人的地方，当我在查询的in子句中给出一个位置列表

我不想找到所有不在人员表中的位置，只有那些我在in子句中提供的位置

查询

SELECT location_id FROM people WHERE location_id in (5,6,7,8,9);

预期输出

|Location_id|
           6|
           9|

Answer 1

使用JOIN代替两个IN操作（提高了性能）。

试试这个：

SELECT pl.location_id
FROM place pl LEFT JOIN
     people pp ON pl.location_id=pp.location_id
WHERE pp.location_id IS NULL 
      AND pl.location_id IN (5,6,7,8,9);

结果：

LOCATION_ID
6
9

请参阅SQL Fiddle中的结果。

修改

不使用JOIN：

SELECT location_id
FROM place
WHERE location_id NOT IN 
          (SELECT location_id FROM people)
AND pl.location_id IN (5,6,7,8,9);

Answer 2

SELECT location_id FROM place WHERE location_id NOT IN (SELECT location_id  FROM people) AND location_id IN (5,6,7,8,9);

Answer 3

尝试以下查询

    select place from place 
          inner join poeple on place.location_id <> people.location_id where AND 
    place.location_id IN (5,6,7,8,9);

Inner join用于加入两个或更多表格。

Answer 4

select p.location_id FROM people p join Place l on 
p.location_id!=l.location_id WHERE p.location_id in (5,6,7,8,9);

Answer 5

从p.location_id = l.location_id中的人们左边的连接位置l中选择l.location_id，其中p.location_id不在（5,6,7,8,9）