Question

我遇到的问题是表格中有多个结果用于我想要查找的内容。如果我有这个：

SELECT DISTINCT student.ident AS [Ident],
     proghist.prgc AS [Test Code]
FROM student 
LEFT OUTER JOIN proghist

对于一些学生，我会得到多个“测试代码”，所以它看起来像这样：

Ident   Test Code
123456   1
123456   4
654321   2
654321   6
122222   1

有没有办法将它们组合成一行和单独的列？

编辑：我希望数据在最终结果中：

123456 1 4 
654321 2 6 
122222 1

Answer 1

由于您使用的是SQL Server，因此如果您使用的是SQL Server 2005+，则可以使用PIVOT功能。如果您知道每TestCodes最多只有ident个select ident, [1] TestCode1, [2] TestCode2 from ( SELECT s.ident AS Ident, p.prgc AS TestCode, row_number() over(partition by s.ident order by p.prgc) rn FROM student s LEFT OUTER JOIN proghist p on s.ident = p.ident ) src pivot ( max(TestCode) for rn in ([1], [2]) ) piv，那么您可以对这些值进行硬编码：

TestCodes

请参阅SQL Fiddle with Demo

如果您有PIVOT的未知数量的值，那么您可以使用动态SQL来DECLARE @cols AS NVARCHAR(MAX), @colsPivot AS NVARCHAR(MAX), @query AS NVARCHAR(MAX) select @cols = STUFF((SELECT distinct ',' + QUOTENAME(rn) FROM ( select cast(row_number() over(partition by s.ident order by p.prgc) as varchar(50)) rn FROM student s LEFT OUTER JOIN proghist p on s.ident = p.ident ) src FOR XML PATH(''), TYPE ).value('.', 'NVARCHAR(MAX)') ,1,1,'') select @colsPivot = STUFF((SELECT distinct ',' + QUOTENAME(rn) + ' as TestCode'+rn FROM ( select cast(row_number() over(partition by s.ident order by p.prgc) as varchar(50)) rn FROM student s LEFT OUTER JOIN proghist p on s.ident = p.ident ) src FOR XML PATH(''), TYPE ).value('.', 'NVARCHAR(MAX)') ,1,1,'') set @query = 'SELECT Ident, ' + @colsPivot + ' from ( SELECT s.ident AS Ident, p.prgc AS TestCode, row_number() over(partition by s.ident order by p.prgc) rn FROM student s LEFT OUTER JOIN proghist p on s.ident = p.ident ) src pivot ( max(TestCode) for rn in (' + @cols + ') ) p ' execute(@query)数据：

PIVOT

请参阅SQL Fiddle with Demo

如果您无权访问CASE函数，则可以使用带select ident, max(case when rn = 1 then testcode else '' end) TestCode1, max(case when rn = 2 then testcode else '' end) TestCode2 from ( SELECT s.ident AS Ident, p.prgc AS TestCode, row_number() over(partition by s.ident order by p.prgc) rn FROM student s LEFT OUTER JOIN proghist p on s.ident = p.ident ) src group by ident语句的聚合函数：

|  IDENT | TESTCODE1 | TESTCODE2 |
----------------------------------
| 122222 |         1 |         0 |
| 123456 |         1 |         4 |
| 654321 |         2 |         6 |

请参阅SQL Fiddle with Demo

这三个都会产生相同的结果：

{{1}}

从同一个表中获取多行

1 个答案: