3 
SELECT 
    [Reg. number], [Surname], 
    [SESREFDATETIME1], [ATTENDANCE1], 
    [SESREFDATETIME2], [ATTENDANCE2],
    [SESREFDATETIME3], [ATTENDANCE3],
    [SESREFDATETIME4], [ATTENDANCE4] 
FROM
    (SELECT
        [Reg. number], [Surname], 
        col + CAST(rn AS varchar(10)) col, 
        value
     FROM
        (SELECT
            [Reg. number], Surname,
            row_number() over(partition by [Reg. number] order by SESREFDATETIME) rn
         FROM #Temp) t
     CROSS APPLY
         (SELECT 'SESREFDATETIME', SESREFDATETIME 
          UNION ALL 
          SELECT 'ATTENDANCE', ATTENDANCE) c (col, value)
    ) x
PIVOT
    (max(value)
     for col in ([SESREFDATETIME1], [ATTENDANCE1], [SESREFDATETIME2],[ATTENDANCE2], [SESREFDATETIME3], [ATTENDANCE3], [SESREFDATETIME4],[ATTENDANCE4])
    ) p;

在我的过程中，我创建了一个#temp临时表，并尝试在多列中显示多行。我使用数据透视的原因是因为这段代码是动态创建的并且行数是未知的。我正在运行代码，但它给出了错误。我要疯了。找不到错误在哪里。它表明在交叉应用中存在无效的列名。我认为还有另一个错误。它显示错误的一面。测试表格式如下

Create table #Temp
(
    [Reg. number] int, 
    [Surname] Varchar(50), 
    SESREFDATETIME Varchar(80), 
    ATTENDANCE Varchar(10)
)

示例数据格式为

2005162 Abasov  04/09/2014 09:00 - 10:00    Y
2005458 Baxşiyev    15/04/2015 01:00 - 04:00    NULL
2005458 Baxşiyev    16/09/2014 14:00 - 17:00    Y
2005538 Abbasbəyli  13/10/2014 12:00 - 15:00    Y
4

1 回答 1

2

您的“交叉应用 x”看不到 SESREFDATETIME 和 ATTENDANCE，如果您在“子选择 t”中包含这些列，则交叉应用部分可以获得值

尝试这个：

SELECT 
    [Reg. number], [Surname], 
    [SESREFDATETIME1], [ATTENDANCE1], 
    [SESREFDATETIME2], [ATTENDANCE2],
    [SESREFDATETIME3], [ATTENDANCE3],
    [SESREFDATETIME4], [ATTENDANCE4] 
FROM
    (SELECT
        [Reg. number], [Surname], 
        col + CAST(rn AS varchar(10)) col, 
        value
     FROM
        (SELECT
            [Reg. number], Surname,
            row_number() over(partition by [Reg. number] order by SESREFDATETIME) rn,
            SESREFDATETIME,
            ATTENDANCE
         FROM #Temp) t
     CROSS APPLY
         (SELECT 'SESREFDATETIME', SESREFDATETIME 
          UNION ALL 
          SELECT 'ATTENDANCE', ATTENDANCE) c (col, value)
    ) x
PIVOT
    (max(value)
     for col in ([SESREFDATETIME1], [ATTENDANCE1], [SESREFDATETIME2],[ATTENDANCE2], [SESREFDATETIME3], [ATTENDANCE3], [SESREFDATETIME4],[ATTENDANCE4])
    ) p;
于 2015-01-29T13:11:50.097 回答