经典sql代码--按年龄段、品牌分类进行分组统计,正在学习分组统计SQL的朋友有福了。
--> 测试数据:[tb] if object_id('[tb]') is not null drop table [tb] go create table [tb]([姓名] varchar(1),[部门] varchar(4),[学历] varchar(4),[出生年月] datetime) insert [tb] select 'A','后勤','高中','1986-1-1' union all select 'B','后勤','初中','1984-3-7' union all select 'C','管理','本科','1987-2-1' union all select 'D','操作','专科','1976-2-1' union all select 'E','操作','专科','1943-2-1' --------------开始查询-------------------------- declare @sql varchar(8000) set @sql = 'select 部门,dbo.AgeLevel([出生年月]) as 年龄段' select @sql = @sql + ' , sum(case 学历 when ''' + 学历 + ''' then 1 else 0 end) [' + 学历 + ']' from (select distinct 学历 from tb) as a set @sql = @sql + ' from tb group by 部门,dbo.AgeLevel([出生年月])' exec(@sql) /* 部门 年龄段 本科 初中 高中 专科 ---- ---------- ----------- ----------- ----------- ----------- 管理 21-30 1 0 0 0 后勤 21-30 0 1 1 0 操作 31-40 0 0 0 1 操作 50以上 0 0 0 1 (4 行受影响) */ drop function AgeLevel go --获取年龄段 create function AgeLevel(@birthday datetime) returns varchar(10) as begin declare @AgeLevel varchar(10) select @AgeLevel=case((datediff(year,@birthday,getdate())-1)/10) when 2 then '21-30' when 3 then '31-40' when 4 then'41-50' else '50以上' end return @AgeLevel end go select * ,dbo.AgeLevel([出生年月]) as 年龄段 from tb /* 姓名 部门 学历 出生年月 年龄段 ---- ---- ---- ----------------------- ---------- A 后勤 高中 1986-01-01 00:00:00.000 21-30 B 后勤 初中 1984-03-07 00:00:00.000 21-30 C 管理 本科 1987-02-01 00:00:00.000 21-30 D 操作 专科 1976-02-01 00:00:00.000 31-40 E 操作 专科 1943-02-01 00:00:00.000 50以上 */ select N'年龄段'=( case((datediff(year,[出生年月],getdate())-1)/10) when 2 then '21-30' when 3 then '31-40' when 4 then'41-50' else '50以上' end), count(*) as count from tb group by ( case((datediff(year,[出生年月],getdate())-1)/10) when 2 then '21-30' when 3 then '31-40' when 4 then'41-50' else '50以上' end ) /* 年龄段 count ------ ----------- 21-30 3 31-40 1 50以上 1 (3 行受影响) */ --以10岁为递增 select cast(f1*10+1 as varchar(3))+'-'+cast(f1*10+10 as varchar(3)) as 年龄段,f2 as 人数 from ( select datediff(d,[出生年月],getdate())/365/10 as f1, count(*) as f2 from tb group by datediff(d,[出生年月],getdate())/365/10) a order by cast(f1*10+1 as varchar(3))+'-'+cast(f1*10+10 as varchar(3)) /* 年龄段 人数 ------- ----------- 21-30 3 31-40 1 61-70 1 (3 行受影响) */ SELECT SUM( CASE WHEN datediff(year, [出生年月], getdate()) BETWEEN 16 AND 20 THEN 1 ELSE 0 END) AS '16-20', SUM(CASE WHEN datediff(year, [出生年月], getdate()) BETWEEN 21 AND 30 THEN 1 ELSE 0 END) AS '21-30', SUM(CASE WHEN datediff(year, [出生年月], getdate()) BETWEEN 31 AND 40 THEN 1 ELSE 0 END) AS '31-40', SUM(CASE WHEN datediff(year, [出生年月], getdate()) BETWEEN 41 AND 50 THEN 1 ELSE 0 END) AS '41-50', SUM(CASE WHEN datediff(year, [出生年月], getdate()) BETWEEN 51 AND 60 THEN 1 ELSE 0 END) AS '51-60', SUM(CASE WHEN datediff(year, [出生年月], getdate()) BETWEEN 61 AND 70 THEN 1 ELSE 0 END) AS '61-70' FROM tb /* 16-20 21-30 31-40 41-50 51-60 61-70 ----------- ----------- ----------- ----------- ----------- ----------- 0 3 1 0 0 1 (1 行受影响) */ create table brands(id int,brand varchar(10), address varchar(10)) insert into brands values(1 ,'联想', '北京') insert into brands values(2 ,'惠普', '美国') insert into brands values(3 ,'神舟', '深圳') create table products(id int, brand int, name varchar(10)) insert into products values(1 ,1, '联想1') insert into products values(2 ,1, '联想2') insert into products values(3 ,2, '惠普1') insert into products values(4 ,2, '惠普2' ) insertinto products values(5 ,1, '联想3') insertinto products values(6 ,3, '神舟1') insertinto products values(7 ,1, '联想4') go select ID=row_number()over(order by getdate()), b.产品数量, a.[brand], a.[address] from brands a, (select [brand], count([brand])产品数量 from products group by [brand] )b where a.[ID]=b.[brand] order by b.产品数量 desc select b.id,b1.cnt as 产品数量,b.brand,b.address from brands b join ( select brand,count(brand) cnt from products group by brand ) b1 on b1.brand=b.id id 产品数量 brand address ----------- ----------- ------------------------------ ------------------------------ 1 4 联想 北京 2 2 惠普 美国 3 1 神舟 深圳 (3 行受影响) select sum(case when ( 字段名>0 and 字段名<4000) then 1 else 0 end) 别名, sum(case when 字段名>=4000 and 字段名<8000 then 1 else 0 end) 别名, sum(case when 字段名>=8000 then 1 else 0 end) 别名 , count(*) as total from 表名
原文链接:http://www.cnblogs.com/zengxiangzhan/archive/2010/02/04/1663468.html