Question

我写了这个查询：

SELECT s, [1] AS a1, [2] AS a2, [3] AS a3, [4] AS a4
FROM (SELECT grade, aid, s FROM m) p
PIVOT
(
SUM(grade)
FOR aid IN ([1], [2], [3], [4])
) AS pvt ORDER BY pvt.s;

返回结果：

s  a1  a2  a3  a4
1  25  69  95  56
2  27  99  16  87
. . . .
99 98  12  34  76

这正是我想要的结果。我的问题是“援助”中并不总是有四个不同的值。是否可以重写此查询（或使用存储过程），以便'a *'列的数量取决于'aid'中有多少个不同的值？

Answer 1

您需要使用Dynamic Pivot来获取所需的列列表。这将首先检索列列表，然后转动该列表。类似的东西：

DECLARE @cols AS NVARCHAR(MAX),
    @query  AS NVARCHAR(MAX);

select @cols = STUFF((SELECT distinct ',' + QUOTENAME(aid) 
            FROM m 
            FOR XML PATH(''), TYPE
            ).value('.', 'NVARCHAR(MAX)') 
        ,1,1,'')

set @query = 'SELECT s, ' + @cols + ' from 
            (
                select grade, aid, s
                from m
           ) x
            pivot 
            (
                sum(grade)
                for aid in (' + @cols + ')
            ) p 
            ORDER BY p.s'

execute(@query)

Answer 2

Lamak：以下是我使用列别名的方法。别名链接到另一个表中由“aid”链接的列中的值。

DECLARE
    @cols AS NVARCHAR(MAX),
    @colsAlias AS NVARCHAR(MAX),
    @query  AS NVARCHAR(MAX);

SELECT @cols = STUFF((SELECT DISTINCT ',' + QUOTENAME(aid) 
    FROM m
    FOR XML PATH(''), TYPE).value('.', 'NVARCHAR(MAX)'),1,1,'')

SELECT @colsAlias = STUFF((SELECT DISTINCT ',' + QUOTENAME(m.aid) + ' AS ' + QUOTENAME(n.aName)  
    FROM m INNER JOIN n ON m.aid = n.aid
    FOR XML PATH(''), TYPE).value('.', 'NVARCHAR(MAX)'),1,1,'')

SET @query = 'SELECT s, ' + @colsAlias + ' FROM 
                (
                SELECT grade, aid, s
                    FROM m
                ) x

            PIVOT 
            (
                MIN(grade) FOR aid IN (' + @cols + ')
            ) p '

EXECUTE(@query)

旋转具有无限量的不同值

2 个答案: