简单sql分组统计问题

[沉默的大多数的杂谈感想]

一个表如下：testdb
name type
a 0
b 0
fd 1
fdf 2
kvlc 1

如何用sql语句统计出type类型相同的记录数，最好在一个sql语句中完成。
多次查询的我也知道。
例如可以统计出： 0值的2条 1值的2条 2值的1条，
type类型个数已经知道为0-5
希望大家给出最有效率的查询

[WangZWang]

--?
Select type,count(*) as cn from tb group by type

[fa_ge]

Select type,count(*) as count from tb group by type

[dawugui]

select type , count(*) cnt from testdb group by type

[dobear_0922]

Select type,Types=count(1) from tb
group by type
order by type

[dawugui]

直接group by 就行了.这分送得有点亏.

[SoftwKLC]

Select Type,[Count]=Count(*) From testdb Group By Type

[dobear_0922]

大哥们，count(1)比count(*)效率高

[love16]

呵呵，这个问题真的有点容易！

[welove1983]

50分那...汗

[tbxl]

declare @t table (name varchar(20),type int)
insert into @t
select 'a', 0
union select 'b', 0
union select 'fd ', 1
union select 'fdf ', 2
union select 'kvlc ', 1

select type ,cn =count(1)
from @t
group by type

[Limpire]

select type, 条=count(*) from testdb group by type

[wuxinyuzhu]

declare @t table(name varchar(20),type int)
insert into @t
select 'a',0
union all select 'b',0
union all select 'fb',1
union all select 'fbf',2
union all select 'fdfs',1


declare @b varchar(1000)
set @b=''
select @b=case @b when ''then '' else @b+',' end +rtrim(name) from @t group by name

select @b name,count_0=sum(case type when 0 then 1 end),
count_1=sum(case type when 1 then 1 end),
count_2=sum(case type when 2 then 1 end)
from @t

/*
是不是要的结果是这样的
name count_0 count_1 count_2
a,b,fb,fbf,fdfs 2 2 1

*/

[pt1314917]

Select type,count(*)[count] from tb group by type
赚了。呵呵``

[wuxinyuzhu]

刚才的错了
改一下
declare @t table(name varchar(20),type int)
insert into @t
select 'a',0
union all select 'b',0
union all select 'fb',1
union all select 'fbf',2
union all select 'fdfs',1


declare @b varchar(1000)
set @b=''
select @b=case @b when ''then '' else @b+',' end +rtrim(name) from @t

select @b name,count_0=sum(case type when 0 then 1 end),
count_1=sum(case type when 1 then 1 end),
count_2=sum(case type when 2 then 1 end)
from @t

[welove1983]

wuxinyuzhu() ... ~~~
这个LZ没要求这么做吧? 汗

[wuxinyuzhu]

是的 是我自己想的 顺便帮我优化一下
请大家指教一下

[woaiguainv]

分析一下:你要得的结果是------统计出type类型相同的记录数所以设计到是TYPE字段

分类查询可以得到：　select type,count(*) as zhonglei from testdb grouop by type


　　　　　　很容易理解，你查询得到的结果是一张表，有type，和zhonglei 两个字段就很容易理解了