高分求其实现方法！值得一看！
有一张表a的数据如下：
NAME NAME COUNT
--------------------------- -------------- -------------
FR 白金 1
FR 宝石 1
FR 黄金 1
FR 其它
[292 byte] By [msdn] at [2007-8-14 15:34:45] 
# 2 Re: 高分求其实现方法！值得一看！
select name,
sum(decode(客户等级,钻石,count,0)) 钻石,
sum(decode(客户等级,翡翠,count,0)) 翡翠,
...
from a
group by name
jiezhi at 2004-5-13 13:09:16 >
# 3 Re: 高分求其实现方法！值得一看！
select name,sum(decode(name_type,'钻石'，type_count，0)) '钻石'，sum(decode(name_type,'翡翠',type_count,0)) '翡翠' from tab_name group by name;
可以通过这个语句进行完善。
zlqs at 2004-5-13 13:12:36 >
# 7 Re: 高分求其实现方法！值得一看！
select Name,sum(case 类型
when ‘白金’ then count end) as "白金",
.......
sum(salary) "Total" from employees
group by name
注意：不能两个字段名都叫name
luoyong777 at 2004-6-9 13:48:53 >