Question

我有FEATURE和FEATURE_DETAILS表。 FEATURE_DETAILS可以有很多功能。

FEATURE_DETAILS
feature_deatails_id | feature_id
1                        1
1                        2
1                        4 
2                        1
2                        2
2                        4
2                        5

我在选择feature_deatails_id时遇到问题5 feature_id。所以结果应该是：

feature_deatails_id | feature_id
1                     null

Answer 1

这应该可以解决你的目的

with data as (
          select 1 as feature_details_id, 1 feature_id from dual
    union select 1,2 from dual
    union select 1,4 from dual
    union select 2,1 from dual
    union select 2,2 from dual
    union select 2,4 from dual
    union select 2,5 from dual
)
select distinct feature_details_id, NULL feature_id from data
where feature_details_id not in (
      select feature_details_id from data
      where  feature_id = 5)
;

上述查询的作用是 -

选择feature_details_id where feature = 5
删除所有feature_details_id并选择剩余的

Answer 2

使用NOT IN应该没问题

SELECT DISTINCT FEATURE_DETAILS_ID, NULL AS FEATURE_ID 
    FROM FEATURE_DETAILS 
    WHERE FEATURE_DETAILS_ID NOT IN (
        SELECT FEATURE_DETAILS_ID FROM FEATURE_DETAILS WHERE FEATURE_ID = 5
    );

Answer 3

选择feature_deatails_id，null为feature_id 来自feature_details feature_id！= 5

这应该可以解决你的目的。

选择不符合条件的行

3 个答案: