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SQL Server 2005以未知列数为轴心

pivot sql-server sql

Mitchel Sellers · 技术社区 · 16 年前

我正在处理一组数据,这些数据如下所示。

StudentName  | AssignmentName |  Grade
---------------------------------------
StudentA     | Assignment 1   | 100
StudentA     | Assignment 2   | 80
StudentA     | Total          | 180
StudentB     | Assignment 1   | 100
StudentB     | Assignment 2   | 80
StudentB     | Assignment 3   | 100
StudentB     | Total          | 280

作业的名称和数量是动态的,我需要得到以下结果。

Student      | Assignment 1  | Assignment 2  | Assignment 3  | Total
--------------------------------------------------------------------
Student A    | 100           | 80            | null          | 180
Student B    | 100           | 80            | 100           | 280

我知道如何使用pivot来完成3个任务,只需简单地命名列,它正试图以一种动态的方式来完成,而我还没有找到一个好的解决方案。我正在尝试在SQLServer2005上执行此操作

编辑

理想情况下,我希望在不使用动态SQL的情况下实现它,因为这违反了策略。如果不可能…那么使用动态SQL的工作示例将起作用。

7 回复 | 直到 16 年前

Community CDub 8 年前

我知道你说没有动力 SQL SQL .

如果你能在我的网站上找到我对类似问题的答案 Pivot Table and Concatenate Columns 和 PIVOT in sql 2005

不存在易受注射伤害的情况,也没有很好的理由禁止注射。另一种可能性(如果数据很少更改)是执行代码生成,而不是动态生成 SQL 是定期生成到存储过程的。

Taryn Frank Pearson 12 年前

到 PIVOT

CREATE TABLE yourtable
    ([StudentName] varchar(8), [AssignmentName] varchar(12), [Grade] int)
;

INSERT INTO yourtable
    ([StudentName], [AssignmentName], [Grade])
VALUES
    ('StudentA', 'Assignment 1', 100),
    ('StudentA', 'Assignment 2', 80),
    ('StudentA', 'Total', 180),
    ('StudentB', 'Assignment 1', 100),
    ('StudentB', 'Assignment 2', 80),
    ('StudentB', 'Assignment 3', 100),
    ('StudentB', 'Total', 280)
;

动态轴:

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

select @cols = STUFF((SELECT distinct ',' + QUOTENAME(AssignmentName) 
                    from yourtable
            FOR XML PATH(''), TYPE
            ).value('.', 'NVARCHAR(MAX)') 
        ,1,1,'')

set @query = 'SELECT StudentName, ' + @cols + ' from 
             (
                select StudentName, AssignmentName, grade
                from yourtable
            ) x
            pivot 
            (
                min(grade)
                for assignmentname in (' + @cols + ')
            ) p '

execute(@query)

看见 SQL Fiddle with Demo

结果是:

| STUDENTNAME | ASSIGNMENT 1 | ASSIGNMENT 2 | ASSIGNMENT 3 | TOTAL |
--------------------------------------------------------------------
|    StudentA |          100 |           80 |       (null) |   180 |
|    StudentB |          100 |           80 |          100 |   280 |

BoltBait 16 年前

tsilb 16 年前

您可以查询信息以获取列名和类型,然后在构建结果集时将结果用作子查询。注意,您可能需要稍微更改登录名的访问权限。

Community CDub 8 年前

这和 PIVOT in sql 2005

http://www.simple-talk.com/community/blogs/andras/archive/2007/09/14/37265.aspx

-1

fancyPants 12 年前

SELECT TrnType
INTO #Temp1
FROM
(
    SELECT '[' + CAST(TransactionType AS VARCHAR(4)) + ']' AS TrnType FROM tblPaymentTransactionTypes
) AS tbl1

SELECT * FROM #Temp1

SELECT * FROM
(
    SELECT FirstName + ' ' + LastName AS Patient, TransactionType, ISNULL(PostedAmount, 0) AS PostedAmount
    FROM tblPaymentTransactions
            INNER JOIN emr_PatientDetails ON tblPaymentTransactions.PracticeID = emr_PatientDetails.PracticeId
            INNER JOIN tblPaymentTransactionDetails ON emr_PatientDetails.PatientId = tblPaymentTransactionDetails.PatientID
                        AND tblPaymentTransactions.TransactionID = tblPaymentTransactionDetails.TransactionID
    WHERE emr_PatientDetails.PracticeID = 152
) tbl
PIVOT (SUM(PostedAmount) FOR [TransactionType] IN (SELECT * FROM #Temp1)
) AS tbl4

-2

Gabriele Petrioli 14 年前

select studentname,[Assign1],[Assign2],[Assign3],[Total] 
from 
(
 select studentname, assignname, grade from student
)s
pivot(sum(Grade) for assignname IN([Assign1],[Assign2],[Assign3],[Total])) as pvt