Sql Pivot水平

Question

下午全部，

试图找出如何在没有聚合的情况下进行旋转，因为存在字母数字值。还有另一个术语吗？

这是我拥有的数据：

Id          Name
1Bravo      John
1Bravo      Smith
2Charlie    Dave
2Charlie    Billy

我希望实现的目标：

Id       Name    Name
1Bravo   John    Smith   

用Bravo显示结果

我还没有通过支点

获得任何好处

Select ID, Name
From Table
Pivot (
Max(ID) For ID Like ('Bravo%') 

感谢任何帮助

由于

Answer 1

这将根据给定Id的名称数量动态构建查询。

--Create Sample Data
CREATE TABLE myTable
    ([Id] varchar(8), [Name] varchar(5))
;
INSERT INTO myTable
    ([Id], [Name])
VALUES
    ('1Bravo', 'John'),('1Bravo', 'Smith'),    
    ('2Charlie', 'Dave'),('2Charlie', 'Billy'),('2Charlie', 'Bobby')
;


DECLARE @Id VARCHAR(8) = '1Bravo',
        @Sql NVARCHAR(MAX),
        @SelCols NVARCHAR(MAX),
        @PivCols NVARCHAR(MAX)

--Build Select and Pivot columns dynamically based on the number of Names for the given @Id
SELECT  @SelCols = COALESCE(@SelCols + ',','') + QUOTENAME(ROW_NUMBER() OVER (ORDER BY [Name])) + ' As [Name]',
        @PivCols = COALESCE(@PivCols + ',','') + QUOTENAME(ROW_NUMBER() OVER (ORDER BY [Name]))
FROM    myTable 
WHERE   [Id] = @Id


--Build the dynamic sql 
SET     @Sql = 'SELECT [Id], ' + @SelCols + ' 
                FROM (  SELECT  [Id], [Name], ROW_NUMBER() OVER (PARTITION BY [Id] ORDER BY [Name]) Rn
                        FROM    myTable 
                        WHERE   [Id] = @Id) mt
                PIVOT (
                    MAX([Name])
                    FOR Rn IN (' + @PivCols + ')) p'

--Execute the dynamic sql passing in the @Id parameter
EXECUTE sp_executesql @Sql, N'@Id VARCHAR(8)', @Id = @Id

--@Id = '1Bravo` result
--Id       Name  Name
---------- ----- -----
--1Bravo   John  Smith

--@Id = '2Charlie` result
--Id       Name  Name  Name
---------- ----- ----- -----
--2Charlie Billy Bobby Dave

Answer 2

您应首先限制您的最大名称数量（我在查询中设置10个名称），然后您可以转动它。

我添加了一个列，其中包含行（cnt）中的名称总数，以便于循环或检查，而无需查找第一个空名称。

;with
YourTable as (
    select * 
    from ( 
    values 
        ('1Bravo',      'John'),
        ('1Bravo',      'Smith'),
        ('2Charlie',    'Dave'),
        ('2Charlie',    'Billy')
    ) t (Id, Name)
),
n as (
    select id, count(distinct name) cnt 
    from YourTable
    group by id
),
x as (
    select *, ROW_NUMBER() over (partition by id order by name) idx
    from YourTable
)
select n.cnt, p.*
from x 
pivot (min(name) for idx in ([1],[2],[3],[4],[5],[6],[7],[8],[9],[10])) p
inner join n on n.Id = p.Id

然后，您可以过滤所需的ID

where n.id like '%bravo%'

我希望这会有所帮助