首先我們建立如下的表：

sc（成績表）
stuid clsid????????????????????????????????????????????? scroe????????????????????????????????????????????????
----- -------- -----------------------------------------------------
0101? 1????????????????????????????????????????????????? 75.0
0102? 1????????????????????????????????????????????????? 70.0
0103? 1????????????????????????????????????????????????? 90.0
0101? 2????????????????????????????????????????????????? 89.0
0102? 2????????????????????????????????????????????????? 80.0
0103? 2????????????????????????????????????????????????? 99.0
0101? 3????????????????????????????????????????????????? 89.0
0102? 3????????????????????????????????????????????????? 79.0
0103? 3????????????????????????????????????????????????? 67.0

其中stuid表示學生編號，clsid表示課程編號，scroe表示成績

stu（學生表）
stuid?????? stuname???????????????????????????????????????????
----------- --------------------------------------------------
101???????? 張三
102???????? 李四
103???????? 王五

cls（課程表）
clsid?????? name??????????????????????????????????????????????
----------- --------------------------------------------------
1?????????? 語文
2?????????? 數學
3?????????? 英語

接下來就是關鍵了，其實也不算好復雜，就是用到了動態sql

declare @sql nvarchar(4000),@sql1 nvarchar(4000)
select @sql='',@sql1=''

select @sql=@sql+',['+name+']=sum(case clsid when '''+clsid+''' then scroe else 0 end)',
?????? @sql1=@sql1+',['+name+'名次]=(select sum(1) from # where ['+name+']>=a.['+name+'])'??????
from(select distinct b.clsid,c.name from sc as b inner join cls as c on c.clsid=b.clsid) as a order by clsid

exec('select stuid 學號'+@sql+',總成績=sum(scroe)
,平均分=Convert(dec(5,1),avg(scroe)),總名次=(select sum(1) from(select stuid,aa=sum(scroe) from sc group by stuid) aa where sum(a.scroe)<=aa) into # from sc as a group by stuid select b.stuname as 姓名,a.*'+@sql1+' from # as a inner join stu as b on a.學號=b.stuid')

以下就是結果：

姓名??? 學號??? 語文??? 數學??? 英語??? 總成績? 平均分 總名次 語文名次 數學名次 英語名次???
---------------------------------------------------- --------------------------------------
張三???? 0101??? 75.0??? 89.0????? 89.0???? 253.0????? 84.3???????? 2???????????? 2??????????????? 2?????????????? 1
李四?????0102??? 70.0??? 80.0????? 79.0???? 229.0????? 76.3???????? 3???????????? 3?????????????? ?3?????????????? 2
王五?????0103??? 90.0??? 99.0????? 67.0???? 256.0????? 85.3???????? 1???????????? 1?????????????? ?1?????????????? 3

這就是交叉表的用法，對于報表統計很有用處，我們可以依此類推寫出功能更強大的查詢。