oracle分析函数

oracle分析函数--SQL*PLUS环境
--1、GROUP BY子句
--CREATE TEST TABLE AND INSERT TEST DATA.
create table students
(id number(15,0),
area varchar2(10),
stu_type varchar2(2),
score number(20,2));

insert into students values(1, '111', 'g', 80 );
insert into students values(1, '111', 'j', 80 );
insert into students values(1, '222', 'g', 89 );
insert into students values(1, '222', 'g', 68 );
insert into students values(2, '111', 'g', 80 );
insert into students values(2, '111', 'j', 70 );
insert into students values(2, '222', 'g', 60 );
insert into students values(2, '222', 'j', 65 );
insert into students values(3, '111', 'g', 75 );
insert into students values(3, '111', 'j', 58 );
insert into students values(3, '222', 'g', 58 );
insert into students values(3, '222', 'j', 90 );
insert into students values(4, '111', 'g', 89 );
insert into students values(4, '111', 'j', 90 );
insert into students values(4, '222', 'g', 90 );
insert into students values(4, '222', 'j', 89 );
commit;

col score format 999999999999.99
--A、GROUPING SETS
select id,area,stu_type,sum(score) score
from students
group by grouping sets((id,area,stu_type),(id,area),id)
order by id,area,stu_type;
/*--------理解grouping sets
select a, b, c, sum( d ) from t
group by grouping sets ( a, b, c )
等效于
select * from (
select a, null, null, sum( d ) from t group by a
union all
select null, b, null, sum( d ) from t group by b
union all
select null, null, c, sum( d ) from t group by c
)
*/
--B、ROLLUP
select id,area,stu_type,sum(score) score
from students
group by rollup(id,area,stu_type)
order by id,area,stu_type;
/*--------理解rollup
select a, b, c, sum( d )
from t
group by rollup(a, b, c);
等效于
select * from (
select a, b, c, sum( d ) from t group by a, b, c
union all
select a, b, null, sum( d ) from t group by a, b
union all
select a, null, null, sum( d ) from t group by a
union all
select null, null, null, sum( d ) from t
)
*/
--C、CUBE
select id,area,stu_type,sum(score) score
from students
group by cube(id,area,stu_type)
order by id,area,stu_type;
/*--------理解cube
select a, b, c, sum( d ) from t
group by cube( a, b, c)
等效于
select a, b, c, sum( d ) from t
group by grouping sets(
( a, b, c ),
( a, b ), ( a ), ( b, c ),
( b ), ( a, c ), ( c ),
() )
*/

--D、GROUPING
/*从上面的结果中我们很容易发现,每个统计数据所对应的行都会出现null,
如何来区分到底是根据那个字段做的汇总呢,grouping函数判断是否合计列!*/
select decode(grouping(id),1,'all id',id) id,
decode(grouping(area),1,'all area',to_char(area)) area,
decode(grouping(stu_type),1,'all_stu_type',stu_type) stu_type,
sum(score) score
from students
group by cube(id,area,stu_type)
order by id,area,stu_type;
--2、OVER()函数的使用
--1、RANK()、DENSE_RANK() 的、ROW_NUMBER()、CUME_DIST()、MAX()、AVG()
break on id skip 1
select id,area,score from students order by id,area,score desc;
select id,rank() over(partition by id order by score desc) rk,score from students;
--允许并列名次、名次不间断
select id,dense_rank() over(partition by id order by score desc) rk,score from students;
--即使SCORE相同，ROW_NUMBER()结果也是不同
select id,row_number() over(partition by ID order by SCORE desc) rn,score from students;
select cume_dist() over(order by id) a, --该组最大row_number/所有记录row_number
row_number() over (order by id) rn,id,area,score from students;
select id,max(score) over(partition by id order by score desc) as mx,score from students;
select id,area,avg(score) over(partition by id order by area) as avg,score from students;

--注意有无order by的区别
--按照ID求AVG
select id,avg(score) over(partition by id order by score desc rows between unbounded preceding
and unbounded following ) as ag,score from students;
--2、SUM()
select id,area,score from students order by id,area,score desc;
select id,area,score,
sum(score) over (order by id,area) 连续求和, --按照OVER后边内容汇总求和
sum(score) over () 总和, -- 此处sum(score) over () 等同于sum(score)
100*round(score/sum(score) over (),4) "份额(%)"
from students;
select id,area,score,
sum(score) over (partition by id order by area ) 连id续求和, --按照id内容汇总求和
sum(score) over (partition by id) id总和, --各id的分数总和
100*round(score/sum(score) over (partition by id),4) "id份额(%)",
sum(score) over () 总和, -- 此处sum(score) over () 等同于sum(score)
100*round(score/sum(score) over (),4) "份额(%)"
from students;
--4、LAG(COL,n,default)、LEAD(OL,n,default) --取前后边N条数据
select id,lag(score,1,0) over(order by id) lg,score from students;
select id,lead(score,1,0) over(order by id) lg,score from students;
--5、FIRST_VALUE()、LAST_VALUE()
select id,first_value(score) over(order by id) fv,score from students;
select id,last_value(score) over(order by id) fv,score from students;
/*而对于last_value() over(order by id),结果是有问题的，因为我们没有按照id分区，所以应该出来的效果应该全部是90（最后一条）。
再看个例子就明白了：
select id,last_value(score) over(order by rownum),score from students;
ID LAST_VALUE(SCORE)OVER(ORDERBYR SCORE
---------------- ------------------------------ ----------------------
1 80 80.00
1 80 80.00
1 89 89.00
1 68 68.00
2 80 80.00
2 70 70.00
2 60 60.00
2 65 65.00
3 75 75.00
3 58 58.00
3 58 58.00
3 90 90.00
4 89 89.00
4 90 90.00
4 90 90.00
4 89 89.00
16 rows selected
当使用last_value分析函数的时候，缺省的WINDOWING范围是RANGE BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW，在进行比较的时候从当前行向前进行比较，所以会出现上边的结果。加上如下的参数，结果就正常了。呵呵。默认窗口范围为所有处理结果。*/
select id,last_value(score) over(order by rownum RANGE BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING),score from students;
/*
ID LAST_VALUE(SCORE)OVER(ORDERBYR SCORE
---------------- ------------------------------ ----------------------
1 89 80.00
1 89 80.00
1 89 89.00
1 89 68.00
2 89 80.00
2 89 70.00
2 89 60.00
2 89 65.00
3 89 75.00
3 89 58.00
3 89 58.00
3 89 90.00
4 89 89.00
4 89 90.00
4 89 90.00
4 89 89.00
16 rows selected
*/
--给出一个例子再次理解分析函数
/*********************************************************************************************

http://www.itpub.net/620932.html
问题提出：
一个高级SQL语句问题
假设有一张表，A和B字段都是NUMBER，
A B
1 2
2 3
3 4
4
有这样一些数据
现在想用一条SQL语句，查询出这样的数据
1－》2－》3—》4
就是说，A和B的数据表示一种连接的关系，现在想通过A的一个值，去查询A所对应的B值，直到B为NULL为止，
不知道这个SQL语句怎么写？请教高手！谢谢
*********************************************************************************************/

参考至：http://zhouwf0726.itpub.net/post/9689/158090

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