Question

下表是REORGANIZED表

TID ITEMS      TIMES  TWU  
1   D           5   633
1   M           5   665
1   R           14  861
1   F           4   871
1   I           8   910
1   A           7   942
1   N           7   950
1   Z           2   986
1   H           2   1020
2   S           4   551
2   R           7   861
2   F           6   871
2   I           4   910
2   A           6   942
2   N           8   950
2   Z           6   986
2   H           2   1020
3   U           4   354
3   V           7   528
3   B           2   641
3   J           4   842
3   F           4   871
3   I           2   910
3   A           6   942
3   N           2   950
3   Z           4   986
4   X           4   338
4   O           2   442
4   D           2   633
4   B           6   641
4   M           1   665
4   F           5   871
4   A           1   942
4   N           7   950
4   Z           10  986
4   H           1   1020
5   T           5   365
5   C           8   370
5   K           7   397
5   Q           5   397
5   P           5   471
5   S           3   551
5   D           1   633
5   B           6   641
5   M           6   665
5   J           4   842
5   R           6   861
5   I           1   910
5   A           4   942
5   Z           10  986
5   H           7   1020    
6   L           5   305
6   U           1   354
6   K           2   397

上表按升序排序。我认为twu的项目的最小值为每个TID的叶子节点。剩余项目是中间节点。下表是LEAFNODES

现在我想选择LEAFNODES中的ITEMS作为重组的中间节点

Answer 1

要获取另一个表中不存在的记录，请使用NOT EXISTS。您想知道是否存在中间记录。您考虑所有记录中没有中间记录的最小记录。因此，所有具有比最小值更高的twu的记录都是中间记录，并且您希望从非中间的叶子节点中选择所有项目。

select *
from leafnodes
where not exists
(
  select *
  from reorganized
  where twu > 
  (
    select min(twu)
    from reorganized leaf
    where leaf.tid = reorganized.tid
  )
  and reorganized.items = leafnodes.items
);

这与IN子句相同，我认为它更具可读性。在这里，您将中间项视为一个集合，并且您希望项目不是该集合。

select *
from leafnodes
where items not in
(
  select items
  from reorganized
  where twu > 
  (
    select min(twu)
    from reorganized leaf
    where leaf.tid = reorganized.tid
  )
);

Answer 2

select * from LEAFNODES

Excluse

(with temp as 
(select TID ,  min(TWU) as TWU from LEAFNODES )

select TID , Items from REORGANISED as a inner join temp  on a.TID=temp.TID and a.TWU=Temp.TWU))

Answer 3

;WITH CTE1 AS
(
    -- Select the TId with minimum value from  REORGANISED
    SELECT TID,MIN(TWU)TWU 
    FROM REORGANISED
    GROUP BY TID
)
,CTE2 AS
(
    -- Now we will get value(ITEMS) for the min value and TID from CTE1 for 
    SELECT C1.TID,R.ITEMS,C1.TWU 
    FROM CTE1 C1 
    JOIN REORGANISED R ON C1.TID=R.TID AND C1.TWU=R.TWU
)
SELECT L.* 
FROM CTE2 C2
JOIN LEAFNODES L ON C2.TID=L.TID AND C2.ITEMS=L.ITEMS

如何在一个表中选择SQL中另一个表中不存在的项

3 个答案: