select  
  name,  
  ltrim(sum(case ok when '成功' then 1 else 0 end)*100.0/count(1))+'%'
from
  tb
group by
  name

------解决方案--------------------
SQL code
select  
  name,
  成功率,
  row_number()over(order by 成功率 desc) as AS
from
( 
select  
  name,  
  ltrim(sum(case ok when '成功' then 1 else 0 end)*100.0/count(1))+'%' as 成功率
from
  tb
group by
  name
)t

------解决方案--------------------
/*
表tbl
ID NAME  OK  
1  张三 成功
2  李四 成功
3  张三 失败
4  黎明 成功
5  刘备 成功
6  李四 失败
如何查出每个人的成功率是多少？ SQL语句。
*/

go
if OBJECT_ID('tbl') is not null
drop table tbl
go
create table tbl(
id varchar(2),
name varchar(20),
ok varchar(4) check (ok in('成功','失败'))
)
go
insert tbl
select '1','张三','成功' union all
select '2','李四','成功' union all
select '3','张三','失败' union all
select '4','黎明','成功' union all
select '5','刘备','失败'

-------------------------------------------------------
select 
 name as 姓名,
 left(成功率,charindex('.',CAST(成功率 as varchar))+2)+'%',
 row_number()over(order by 成功率 desc) as 排名
from
(  
select 
 name, 
 ltrim(sum(case ok when '成功' then 1 else 0 end)*100.0/count(1))+'%' as 成功率
from
 tbl
group by
 name
)t

------解决方案--------------------
SQL code
create table tb(id int,name nvarchar(20),ok nvarchar(4))
go
insert tb
select 1,'张三','成功' union all
select 2,'李四','成功' union all
select 3,'张三','失败' union all
select 4,'黎明','成功' union all
select 5,'刘备','失败'
go
select name as 姓名,LTRIM(s)+'%' as 成功率,RANK()over(order by s desc)排名 from(
select name,SUM(case when ok='成功' then 100 else 0 end)/COUNT(*)s from tb group by name
)t
/*
姓名                   成功率           排名
-------------------- ------------- --------------------
黎明                   100%          1
李四                   100%          1
张三                   50%           3
刘备                   0%            4

(4 行受影响)

*/
go
drop table tb