Question

我已经定义了以下三个表：

table 'A':
-------------------
majorID    | bigint (primary key)
-------------------

table 'B':
-------------------
majorID    | bigint (foreign key to table 'A's majorID)
minorID    | bigint (primary key)
totalSize  | bigint
-------------------

table 'C':
-------------------
objectID   | bigint (primary key)
majorID    | bigint (foreign key to table 'A's majorID)
minorID    | bigint (foreign key to table 'B's minorID)
startPoint | bigint
length     | bigint
-------------------

我要做的是获取表'B'中所有行的列表，但显示每行剩余的空间。

可以通过找到最高的“startPoint”找到剩余空间，为包含最高“startPoint”的行添加“length”列的值，然后从表中的“totalSize”列中减去该组合值B'

我目前能够使用以下代码实现此目的：

create table #results (MinorID bigint, MajorID bigint, RemainingSpace bigint)

DECLARE @MinorID bigint
DECLARE @TotalSpace bigint
DECLARE @MajorID bigint
DECLARE cur CURSOR FOR
SELECT MinorID, MajorID, TotalSize FROM B
OPEN cur


FETCH NEXT FROM cur INTO @MinorID,@MajorID, @TotalSpace

WHILE @@FETCH_STATUS = 0
BEGIN
    DECLARE @UsedSize bigint

    SELECT TOP 1 @UsedSize = StartPoint + [length] FROM C
    WHERE MinorID = @MinorID AND MajorID = @MajorID
    ORDER BY StartPoint DESC

    INSERT INTO #results VALUES (@MinorID,@MajorID,@TotalSpace - @UsedSize)

    FETCH NEXT FROM cur INTO @MinorID,@MajorID, @TotalSpace
END

CLOSE cur
DEALLOCATE cur

SELECT * FROM #results
drop table #results

问题是我希望这些表变得非常大，我意识到在表上运行光标可能不是实现我想要的最快方法。

然而，我正在努力寻找更好的解决方案（星期一早上布鲁斯），并希望有人在SQL上更清醒/更好，而不是我可以提出解决方案！

注意：表格设计不是“一成不变”的，所以如果唯一的解决方案是对数据进行非规范化，以便表'B'记录它的“占用空间”，那么我就是开放的那...

编辑：

我使用了已接受答案的修改版本，如下所示：

SELECT B.*, coalesce(C.StartPoint + C.Length,0) AS UsedSize
    FROM TableB B
    LEFT JOIN 
    (
      SELECT *, DENSE_RANK() OVER(PARTITION BY C.MajorID, C.MinorID ORDER BY C.StartPoint DESC) AS Rank
      FROM TableC C
    ) C
    ON C.MajorID = B.MajorID
    AND C.MinorID = B.MinorID
    AND C.Rank = 1

Answer 1

也许您可以使用DENSE_RANK。

在此查询中，我将使用额外列Rank加入表C.如果该列的最高起点，则该列的值为1。在（AND C.Rank = 1）中，我们只提取该行。

SELECT B.*, (C.StartPoint + C.Length) AS UsedSize
FROM TableB B
INNER JOIN 
(
  SELECT *, DENSE_RANK() OVER(PARTITION BY C.MajorID, C.MinorID ORDER BY C.StartPoint DESC) AS Rank
  FROM TableC C
) C
ON C.MajorID = B.MajorID
AND C.MinorID = B.MinorID
AND C.Rank = 1

Answer 2

WITH UsedSpace AS
(
SELECT minorID, MAX(startPoint + length) AS used
FROM C
GROUP BY minorID
)
SELECT B.minorID, totalSize - COALESCE(UsedSpace.used, 0)
FROM B LEFT JOIN UsedSpace ON B.minorID = UsedSpace.minorID

Answer 3

也许你让事情变得更复杂。您正在寻找每个minorID的最大startPoint，以便增加长度并获得使用的大小。但是，它的长度是如此之大以至于加上两者都会超过最大startPoint加上它的长度，它是否有可能有一个较小的startPoint？

这是否可能（max（startPoint）低于其他某个startPoint +长度）：

minorID startPoint length
1       1          10
1       9          3

如果没有，我假设你可以简单地减去max（startPoint + length）：

select 
  minorID, 
  totalSize, 
  totalSize - (select max(startPoint + length) from C where C.minorID = B.minorID) as space_left
from B;

编辑：我刚看了你的评论，有时B没有C存在。为了解决这个问题，你将不得不使用ISNULL或COALESCE：

select 
  minorID, 
  totalSize, 
  totalSize - coalesce((select max(startPoint + length) from C where C.minorID = B.minorID), 0) as space_left
from B;

Answer 4

您可以使用OUTER APPLY从StartPoint

订购的C中获取最高记录

SELECT  B.MajorID,
        B.MinorID,
        B.TotalSize,
        C.StartPoint,
        C.Length,
        SpaceRemaining = B.TotalSize - ISNULL(C.StartPoint + C.Length, 0)
FROM    B
        OUTER APPLY
        (   SELECT  TOP 1 C.StartPoint, C.Length
            FROM    C
            WHERE   B.MinorID = c.MinorID
            ORDER BY C.StartPoint DESC
        ) C;

或者你可以使用ROW_NUMBER来获得相同的结果，这取决于索引等，一个可能比另一个表现更好：

SELECT  B.MajorID,
        B.MinorID,
        B.TotalSize,
        C.StartPoint,
        C.Length,
        SpaceRemaining = B.TotalSize - ISNULL(C.StartPoint + C.Length, 0)
FROM    B
        LEFT JOIN 
        (   SELECT  C.MinorID,
                    C.StartPoint, 
                    C.Length,
                    RowNumber = ROW_NUMBER() OVER(PARTITION BY C.MinorID ORDER BY C.StartPoint DESC, Length DESC)
            FROM    C
        ) C
            ON B.MinorID = c.MinorID
            AND C.Rownumber = 1;

<强> Examples on SQL Fiddle

Answer 5

SELECT B.MajorId, B.MinorId, B.totalSize-(C.length+C.startPoint) as Space 
from TABLEB B 
LEFT JOIN (SELECT MAX(startPoint) maxSP,majorid, minorid FROM TABLEC GROUP BY MajorId, MinorId) 
mxT ON B.majorID  = mxT.majorID AND B.minorId=mxt.minorId 
LEFT JOIN TABLEC C  on C.majorid=mxt.MajorId AND C.minorId=mxt>MinorId AND C.startPoint=mxT.maxSP

从SQL中获取单个MAX值+另一个列值的最佳方法

5 个答案: