如何实现横向聚合

问题描述：

有表tb，数据如下

A<chmetcnv w:st="on" tcsc="0" numbertype="1" negative="False" hasspace="True" sourcevalue="1" unitname="a">1 A</chmetcnv><chmetcnv w:st="on" tcsc="0" numbertype="1" negative="False" hasspace="True" sourcevalue="2" unitname="a">2 A</chmetcnv><chmetcnv w:st="on" tcsc="0" numbertype="1" negative="False" hasspace="True" sourcevalue="3" unitname="a">3 A</chmetcnv><chmetcnv w:st="on" tcsc="0" numbertype="1" negative="False" hasspace="True" sourcevalue="4" unitname="a">4 A</chmetcnv>5

1 2 5 3 4

2 2 3 4 5

0 3 4 2 5

如何输出

A<chmetcnv w:st="on" tcsc="0" numbertype="1" negative="False" hasspace="True" sourcevalue="1" unitname="a">1 A</chmetcnv><chmetcnv w:st="on" tcsc="0" numbertype="1" negative="False" hasspace="True" sourcevalue="2" unitname="a">2 A</chmetcnv><chmetcnv w:st="on" tcsc="0" numbertype="1" negative="False" hasspace="True" sourcevalue="3" unitname="a">3 A</chmetcnv><chmetcnv w:st="on" tcsc="0" numbertype="1" negative="False" hasspace="True" sourcevalue="4" unitname="a">4 A</chmetcnv>5 最大 最小 5以上个数

1 2 5 3 4 5 1 1

2 2 3 4 5 5 2 1

0 3 5 2 6 6 0 2

答：

SQL Server的聚合函数是在列上的，所以可以写自定义的函数，也可以想办法把列变为行，再用聚合函数处理

解决示例

--测试数据

DECLARE @t TABLE(A1 int, A2 int, A3 int, A4 int, A5 int)

INSERT @t SELECT 1,2,5,3,4

<place w:st="on"><span lang="EN-US" style='FONT-SIZE: 9pt; COLOR: blue; FONT-FAMILY: "Courier New"; mso-font-kerning: 0pt; mso-no-proof: yes'>UNION</span></place> ALL SELECT 2,2,3,4,5

<place w:st="on"><span lang="EN-US" style='FONT-SIZE: 9pt; COLOR: blue; FONT-FAMILY: "Courier New"; mso-font-kerning: 0pt; mso-no-proof: yes'>UNION</span></place> ALL SELECT 0,3,4,2,5

--查询

SELECT *,

[min] = (

SELECT MIN(v) FROM(

SELECT v=A.A1 UNION SELECT v=A.A2 UNION SELECT v=A.A3

UNION SELECT v=A.A4 UNION SELECT v=A.A5

)B),

[max] = (

SELECT MAX(v) FROM(

SELECT v=A.A1 UNION SELECT v=A.A2 UNION SELECT v=A.A3

UNION SELECT v=A.A4 UNION SELECT v=A.A5

)B),

[count>=5] = (

SELECT COUNT(*) FROM(

SELECT v=A.A1 UNION SELECT v=A.A2 UNION SELECT v=A.A3

UNION SELECT v=A.A4 UNION SELECT v=A.A5

)B WHERE v>=5)

FROM @t A

查询结果:

A<chmetcnv w:st="on" tcsc="0" numbertype="1" negative="False" hasspace="True" sourcevalue="1" unitname="a">1<span style="mso-spacerun: yes"> </span>A</chmetcnv><chmetcnv w:st="on" tcsc="0" numbertype="1" negative="False" hasspace="True" sourcevalue="2" unitname="a">2<span style="mso-spacerun: yes"> </span>A</chmetcnv><chmetcnv w:st="on" tcsc="0" numbertype="1" negative="False" hasspace="True" sourcevalue="3" unitname="a">3<span style="mso-spacerun: yes"> </span>A</chmetcnv><chmetcnv w:st="on" tcsc="0" numbertype="1" negative="False" hasspace="True" sourcevalue="4" unitname="a">4<span style="mso-spacerun: yes"> </span>A</chmetcnv>5 min max count>=5

----------- ----------- ----------- ----------- ----------- ----------- -----------

1 2 5 3 4 1 5 1

2 2 3 4 5 2 5 1

0 3 4 2 5 0 5 1