鉴于以下表格:
表格对象
id Name rating
1 Megan 9
2 Irina 10
3 Vanessa 7
4 Samantha 9
5 Roxanne 1
6 Sonia 8
交换表
id swap_proposalid counterpartyid
1 4 2
2 3 2
每个人都想要十个。我想为Irina列出可能的掉期列表,其中id 4和3没有出现,因为命题已经存在。
OUTPUT1
id Name rating
1 Megan 9
5 Roxanne 1
6 Sonia 8
由于
答案 0 :(得分:3)
这应该可以解决问题:
SELECT o.id, o.Name, o.rating
FROM objects o
LEFT JOIN swap s on o.id = s.swap_proposalid
WHERE s.id IS NULL
AND o.Name != 'Irina'
答案 1 :(得分:0)
这有效
SELECT mt2.ID, mt2.Name, mt2.Rating
FROM [MyTable] mt2 -- Other Candidates
, [MyTable] mt1 -- Candidate / Subject (Irina)
WHERE mt2.ID NOT IN
(
SELECT st.swap_proposalid
FROM SwapTable st
WHERE
st.counterpartyid = mt1.ID
)
AND mt1.ID <> mt2.ID -- Don't match Irina with Irina
AND mt1.Name = 'Irina' -- Find other swaps for Irina
- 测试数据
CREATE TABLE MyTable
(
ID INT,
Name VARCHAR(100),
Rating INT
)
GO
CREATE TABLE SwapTable
(
ID INT,
swap_proposalid INT,
counterpartyid INT
)
GO
INSERT INTO MyTable VALUES(1 ,'Megan', 9)
INSERT INTO MyTable VALUES(2 ,'Irina', 10)
INSERT INTO MyTable VALUES(3 ,'Vanessa', 7)
INSERT INTO MyTable VALUES(4 ,'Samantha', 9)
INSERT INTO MyTable VALUES(5 ,'Roxanne', 1)
INSERT INTO MyTable VALUES(6 ,'Sonia', 8)
INSERT INTO SwapTable(ID, swap_proposalid, counterpartyid)
VALUES (1, 4, 2)
INSERT INTO SwapTable(ID, swap_proposalid, counterpartyid)
VALUES (1, 3, 2)
答案 2 :(得分:0)
猜测逻辑涉及识别对象,除了评级最高的对象外,具有最高评级对象的命题,例如(使用样本DDL和@nonnb发布的数据):
WITH ObjectHighestRated
AS
(
SELECT ID
FROM MyTable
WHERE Rating = (
SELECT MAX(T.Rating)
FROM MyTable T
)
),
PropositionsForHighestRated
AS
(
SELECT swap_proposalid AS ID
FROM SwapTable
WHERE counterpartyid IN (SELECT ID FROM ObjectHighestRated)
),
CandidateSwappersForHighestRated
AS
(
SELECT ID
FROM MyTable
EXCEPT
SELECT ID
FROM ObjectHighestRated
EXCEPT
SELECT ID
FROM PropositionsForHighestRated
)
SELECT *
FROM MyTable
WHERE ID IN (SELECT ID FROM CandidateSwappersForHighestRated);