Question

我有两张桌子

程序
学生

学生：

Name    Status1   Syear SCode
--------------------
kk       A       2000 1
ra       A       2001  2
Paras    L       2000  2
Prit     L       2001  2
Poot     A       2002  4

程序：

PName       PCode
--------------------
Msc          1
DC           2
PO           4

加入ID

必需的输出：

SELECT * 
FROM 
     (SELECT  
         Program.PName AS v, Status1
      FROM 
         Student, Program  
      WHERE 
         Student.PCode = SCode
      GROUP BY  
         Program.PName, Student.Syear, Status1) AS src
pivot
(
  count(v)
  FOR Status1 IN ([A],)
) as piv

它不会在输出中显示PNAME

          A  L
        -----------
          1  0
          1  2
          1  0

期望的输出

    PNAME  A  L
    -----------
    Msc   1  0
    DC    1  2
    PO    1  0

Answer 1

<强> 1。 STATIC PIVOT

如果事先知道列名

，则可以执行此操作

SELECT PName,ISNULL([A],0) [A],ISNULL([L],0)[L] FROM 
 (
    -- Source data for pivoting
    SELECT P.PName,Status1,
    COUNT(Status1)OVER(PARTITION BY PNAME,Status1)CNT
    FROM #PROGRAM P
    JOIN #Student S ON P.PCODE=S.SCODE
 ) x
 PIVOT 
 (
     --Defines the values in each dynamic columns
     MIN(CNT)
     -- Get the names of columns to pivot
     FOR Status1 IN ([A],[L])
) p            
ORDER BY PName

Click here查看结果

<强> 2。动态枢纽

如果事先不知道列数，可以进行动态旋转。

首先，动态获取列以进行转动

DECLARE @cols NVARCHAR (MAX)

SELECT @cols = COALESCE (@cols + ',[' + Status1 + ']', '[' + Status1 + ']')
               FROM    (SELECT DISTINCT Status1 FROM #Student) PV 
               ORDER BY Status1

现在，下面的变量用于将NULL替换为零。

 DECLARE @NulltoZeroCols NVARCHAR (MAX)

SELECT @NullToZeroCols = SUBSTRING((SELECT ',ISNULL(['+Status1+'],0) AS ['+Status1+']' 
FROM (SELECT DISTINCT Status1 FROM #Student)TAB  
ORDER BY Status1 FOR XML PATH('')),2,8000)

现在转动它。我写了

里面的逻辑

DECLARE @query NVARCHAR(MAX)
SET @query = '-- This outer query forms your pivoted result
             SELECT PName,'+@NullToZeroCols+' FROM 
             (
                -- Source data for pivoting
                SELECT P.PName,Status1,
                COUNT(Status1)OVER(PARTITION BY PNAME,Status1)CNT
                FROM #PROGRAM P
                JOIN #Student S ON P.PCODE=S.SCODE
             ) x
             PIVOT 
             (
                 --Defines the values in each dynamic columns
                 MIN(CNT)
                 -- Get the names from the @cols variable to show as column
                 FOR Status1 IN (' + @cols + ')
            ) p            
            ORDER BY PName;' 

EXEC SP_EXECUTESQL @query

Click here查看结果

将列值（行）转换为列，并在两个不同表的两行上聚合计数

1 个答案: