问一个分析函数的有关问题，如果在分析函数中加入条件

http://www.itpub.net/viewthread.php?tid=608107&extra=&page=1oracle分析函数oracle分析函数zhouwf0726 | 25 七月, 2006 12:51oracle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 SETSselect id,area,stu_type,sum(score) score from studentsgroup by grouping sets((id,area,stu_type),(id,area),id)order by id,area,stu_type;/*--------理解grouping setsselect a, b, c, sum( d ) from tgroup by grouping sets ( a, b, c )等效于select * from (select a, null, null, sum( d ) from t group by aunion allselect null, b, null, sum( d ) from t group by b union allselect null, null, c, sum( d ) from t group by c )*/--B、ROLLUPselect id,area,stu_type,sum(score) score from studentsgroup by rollup(id,area,stu_type)order by id,area,stu_type;/*--------理解rollupselect a, b, c, sum( d )from tgroup by rollup(a, b, c);等效于select * from (select a, b, c, sum( d ) from t group by a, b, c union allselect a, b, null, sum( d ) from t group by a, bunion allselect a, null, null, sum( d ) from t group by aunion allselect null, null, null, sum( d ) from t)*/--C、CUBEselect id,area,stu_type,sum(score) score from studentsgroup by cube(id,area,stu_type)order by id,area,stu_type;/*--------理解cubeselect a, b, c, sum( d ) from tgroup by cube( a, b, c)等效于select a, b, c, sum( d ) from tgroup 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) scorefrom studentsgroup 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 1select 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AVGselect 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;http://zhouwf0726.itpub.net/post/9689/158090-------------------------------------------------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 B1 22 33 44 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*********************************************************************************************/--以下是利用分析函数的一个简单解答：--start with connect by可以参考http://www.itpub.net/620427.htmlCREATE TABLE TEST(COL1 NUMBER(18,0),COL2 NUMBER(18,0));INSERT INTO TEST VALUES(1,2);INSERT INTO TEST VALUES(2,3);INSERT INTO TEST VALUES(3,4);INSERT INTO TEST VALUES(4,NULL);INSERT INTO TEST VALUES(5,6);INSERT INTO TEST VALUES(6,7);INSERT INTO TEST VALUES(7,8);INSERT INTO TEST VALUES(8,NULL);INSERT INTO TEST VALUES(9,10);INSERT INTO TEST VALUES(10,NULL);INSERT INTO TEST VALUES(11,12);INSERT INTO TEST VALUES(12,13);INSERT INTO TEST VALUES(13,14);INSERT INTO TEST VALUES(14,NULL);CREATE TABLE TEST(COL1 NUMBER(18,0),COL2 NUMBER(18,0));INSERT INTO TEST VALUES(1,2);INSERT INTO TEST VALUES(2,3);INSERT INTO TEST VALUES(3,4);INSERT INTO TEST VALUES(4,NULL);INSERT INTO TEST VALUES(5,6);INSERT INTO TEST VALUES(6,7);INSERT INTO TEST VALUES(7,8);INSERT INTO TEST VALUES(8,NULL);INSERT INTO TEST VALUES(9,10);INSERT INTO TEST VALUES(10,NULL);select max(col) from(select SUBSTR(col,1,CASE WHEN INSTR(col,'->')>0 THEN INSTR(col,'->') - 1 ELSE LENGTH(col) END) FLAG,col from(select ltrim(sys_connect_by_path(col1,'->'),'->') col from (select col1,col2,CASE WHEN LAG(COL2,1,NULL) OVER(ORDER BY ROWNUM) IS NULL THEN 1 ELSE 0 END FLAG from test)start with flag=1 connect by col1=prior col2))group by flag;