group by 分组时 or 的有关问题，疑难杂症，坐等牛人。_Oracle技术

group by 分组时 or 的问题，疑难杂症，坐等牛人。。

SQL code

create  table ttt         (          id number(4) primary key,          name varchar2(16) ,          sex varchar2(2),          age number(4)         )                  insert into ttt values (1, 'a','男', 2);         insert into ttt values (6, 'a','男', 4);         insert into ttt values (3, 'a','男', 3);         insert into ttt values (8, 'a','女', 4);                  insert into ttt values (7, 'b','女', 4);         insert into ttt values (2, 'b','女', 2);         insert into ttt values (9, 'b','女', 4);                  insert into ttt values (4, 'c','女', 4);         insert into ttt values (5, 'e','女', 5);

要求：统计数据库中重复数据的条数（有可能不止重复两条）
重复条件：(name相同，sex相同) 或者 (name相同，age相同)，去掉统计时的重复记录（如name相同，sex相同，age相同的记录，不能重复统计）
正确答案：7条
求用sql查询实现
已知算法：
1.

SQL code

--这个算法最简单，但是真实数据量很大，这个效率太差select count(distinct t.id) from ttt t,         (select name,0 age,sex from ttt group by name,sex having count(1)>1 union         select name,age,'' sex from ttt group by name,age having count(1)>1) m where          t.name=m.name and (t.age=m.age or t.sex=m.sex);

SQL code

--此种方式不能去除重复记录，查出来为10条select sum(count(1)) coun from ttt group by grouping sets((name,sex),(name,age)) having count(1)>1;

等sql牛人来解决。。。。。。

------解决方案--------------------
--查出所有重复记录
select name,sex,age,id from(
select name,sex,age,id from ttt where exists (select name,sex from
(select name,sex,count(name) from ttt group by name,sex having count(name)>=2)a
where ttt.name=a.name and ttt.sex=a.sex))a
union
( select name,sex,age,id from ttt where exists (select name,age from
(select name,age,count(name) from ttt group by name,age having count(name)>=2)a
where ttt.name=a.name and ttt.age=a.age))
------解决方案--------------------
--查询重复条数
select count(*) from (
select name,sex,age,id from(
select name,sex,age,id from ttt where exists (select name,sex from
(select name,sex,count(name) from ttt group by name,sex having count(name)>=2)a
where ttt.name=a.name and ttt.sex=a.sex))a
union
( select name,sex,age,id from ttt where exists (select name,age from
(select name,age,count(name) from ttt group by name,age having count(name)>=2)a
where ttt.name=a.name and ttt.age=a.age))
)

------解决方案--------------------
Have a try.

SQL code

with a as   (select id, count(*) over(partition by t.name, t.sex) count from ttt t),  b as   (select id, count(*) over(partition by t.name, t.age) count from ttt t)  select count(id)    from (select a.id            from a           where a.count >= 2          union          select b.id from b where b.count >= 2)
------解决方案--------------------
在原来基础上还可以继续优化，代码如下：
SQL codeselect count(*)  from (select id,               count(*) over(partition by t.name, t.sex) count1,               count(*) over(partition by t.name, t.age) count2          from ttt t) a where a.count1 > 1    or a.count2 > 1
------解决方案--------------------
探讨

引用:

在原来基础上还可以继续优化，代码如下：
SQL code
select count(*)