仅从 mariadb 中获取完全匹配的行 table

Fetching Exact matching rows only From mariadb table

我有一个 table 如下所示

Create Table tmp_test (
    id int(11) unsigned Auto_increment primary key , 
    contract_id int(11) unsigned, 
    item_id int(11) unsigned
);

Insert Into tmp_test (contract_id,item_id)
Values (10,1),(10,2),(10,3),(12,1),(12,2),(14,1),
(16,1),(16,2),(16,3),(16,4),(18,1),(18,2),(20,1),
(20,2),(20,3),(22,2),(22,3),(22,4),(24,1),(24,4);

当我们select查询

1。 Select 不同 contract_id 来自 tmp_test 其中 FIND_IN_SET(item_id, '1,2');

输出需要

12, 18
  1.   Select Distinct contract_id 
      From tmp_test 
      Where FIND_IN_SET(item_id, '1'); 
    

输出需要

14
  1.   Select Distinct contract_id 
      From tmp_test 
      Where FIND_IN_SET(item_id, '1,2,3'); 
    

输出需要

10, 20
  1.   Select Distinct contract_id 
      From tmp_test 
      Where FIND_IN_SET(item_id, '4,1'); 
    

输出需要

24

请帮助我在单个查询中实现这个

此致,

费萨尔

您希望 contract_id 具有输入列表的所有值,而没有其他值。

您可以使用聚合和 having 子句来表达这一点:

select contract_id
from tmp_test
group by item_id
having sum(item_id in (1, 2)) = 2 and sum(item_id not in (1, 2)) = 0

或者如果你更喜欢find_in_set()

having sum(find_in_set(item_id, '1,2')) = 2 and sum(find_in_set(item_id, '1,2') = 0) = 0
Select contract_id From (
Select contract_id,group_concat(item_id order By item_id) items From tmp_test 
Group By contract_id  ) sb
Where items Like '1,2';