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三、常见分析函数详解

为了方便进行实践,特将演示表和数据罗列如下:

一、创建表

create table t( 
bill_month varchar2(12) ,
area_code number,
net_type varchar(2),
local_fare number
);

      

二、插入数据

insert into t values('200405',5761,'G', 7393344.04); 
insert into t values('200405',5761,'J', 5667089.85);
insert into t values('200405',5762,'G', 6315075.96);
insert into t values('200405',5762,'J', 6328716.15);
insert into t values('200405',5763,'G', 8861742.59);
insert into t values('200405',5763,'J', 7788036.32);
insert into t values('200405',5764,'G', 6028670.45);
insert into t values('200405',5764,'J', 6459121.49);
insert into t values('200405',5765,'G', 13156065.77);
insert into t values('200405',5765,'J', 11901671.70);
insert into t values('200406',5761,'G', 7614587.96);
insert into t values('200406',5761,'J', 5704343.05);
insert into t values('200406',5762,'G', 6556992.60);
insert into t values('200406',5762,'J', 6238068.05);
insert into t values('200406',5763,'G', 9130055.46);
insert into t values('200406',5763,'J', 7990460.25);
insert into t values('200406',5764,'G', 6387706.01);
insert into t values('200406',5764,'J', 6907481.66);
insert into t values('200406',5765,'G', 13562968.81);
insert into t values('200406',5765,'J', 12495492.50);
insert into t values('200407',5761,'G', 7987050.65);
insert into t values('200407',5761,'J', 5723215.28);
insert into t values('200407',5762,'G', 6833096.68);
insert into t values('200407',5762,'J', 6391201.44);
insert into t values('200407',5763,'G', 9410815.91);
insert into t values('200407',5763,'J', 8076677.41);
insert into t values('200407',5764,'G', 6456433.23);
insert into t values('200407',5764,'J', 6987660.53);
insert into t values('200407',5765,'G', 14000101.20);
insert into t values('200407',5765,'J', 12301780.20);
insert into t values('200408',5761,'G', 8085170.84);
insert into t values('200408',5761,'J', 6050611.37);
insert into t values('200408',5762,'G', 6854584.22);
insert into t values('200408',5762,'J', 6521884.50);
insert into t values('200408',5763,'G', 9468707.65);
insert into t values('200408',5763,'J', 8460049.43);
insert into t values('200408',5764,'G', 6587559.23);
insert into t values('200408',5764,'J', 7342135.86);
insert into t values('200408',5765,'G', 14450586.63);
insert into t values('200408',5765,'J', 12680052.38);
commit;

            

三、first_value()与last_value():求最值对应的其他属性
问题、取出每月通话费最高和最低的两个地区。

SELECT BILL_MONTH, 
AREA_CODE,
SUM(LOCAL_FARE) LOCAL_FARE,
FIRST_VALUE(AREA_CODE)
OVER(PARTITION BY BILL_MONTH
ORDER BY SUM(LOCAL_FARE) DESC
ROWS BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING) FIRSTVAL,
LAST_VALUE(AREA_CODE)
OVER(PARTITION BY BILL_MONTH
ORDER BY SUM(LOCAL_FARE) DESC
ROWS BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING) LASTVAL
FROM T
GROUP BY BILL_MONTH, AREA_CODE
ORDER BY BILL_MONTH

运行结果:

   

四、rank(),dense_rank()与row_number():求排序

rank,dense_rank,row_number函数为每条记录产生一个从1开始至n的自然数,n的值可能小于等于记录的总数。这3个函数的唯一区别在于当碰到相同数据时的排名策略
①row_number:
row_number函数返回一个唯一的值,当碰到相同数据时,排名按照记录集中记录的顺序依次递增
②dense_rank:
dense_rank函数返回一个唯一的值,当碰到相同数据时,此时所有相同数据的排名都是一样的
③rank:
rank函数返回一个唯一的值,当碰到相同的数据时,此时所有相同数据的排名是一样的,同时会在最后一条相同记录和下一条不同记录的排名之间空出排名

          

演示数据在Oracle自带的scott用户下:
1、rank()值相同时排名相同,其后排名跳跃不连续

SELECT * 
FROM (SELECT DEPTNO,
RANK() OVER(PARTITION BY DEPTNO ORDER BY SAL DESC) RW,
ENAME,
SAL
FROM SCOTT.EMP)
WHERE RW <= 4;

运行结果:


2、dense_rank()值相同时排名相同,其后排名连续不跳跃

SELECT * 
FROM (SELECT DEPTNO,
DENSE_RANK() OVER(PARTITION BY DEPTNO ORDER BY SAL DESC) RW,
ENAME,
SAL
FROM SCOTT.EMP)
WHERE RW <= 4;

运行结果:


3、row_number()值相同时排名不相等,其后排名连续不跳跃

SELECT * 
FROM (SELECT DEPTNO,
ROW_NUMBER() OVER(PARTITION BY DEPTNO ORDER BY SAL DESC) RW,
ENAME,
SAL
FROM SCOTT.EMP)
WHERE RW <= 4;

运行结果:

 

五、lag()与lead():求之前或之后的第N行
lag和lead函数可以在一次查询中取出同一字段的前n行的数据和后n行的值。这种操作可以使用对相同表的表连接来实现,不过使用lag和lead有更高的效率。
lag(arg1,arg2,arg3)
第一个参数是列名,
第二个参数是偏移的offset,
第三个参数是超出记录窗口时的默认值。
  
举例如下:
SQL> select *  from kkk;                                         
                                                                 
        ID NAME                                                  
---------- --------------------                                  
         1 1name                                                 
         2 2name                                                 
         3 3name                                                 
         4 4name                                                 
         5 5name                                                 
SQL> select id,name,lag(name,1,0) over(order by id) from kkk;
                                                                 
        ID NAME                 LAG(NAME,1,0)OVER(ORDERBYID)     
---------- -------------------- ----------------------------     
         1 1name                0                                
         2 2name                1name                            
         3 3name                2name                            
         4 4name                3name                            
         5 5name                4name

SQL> select id,name,lead(name,1,0) over(order by id) from kkk;
                                                                 
        ID NAME                 LEAD(NAME,1,0)OVER(ORDERBYID)    
---------- -------------------- -----------------------------    
         1 1name                2name                            
         2 2name                3name                            
         3 3name                4name                            
         4 4name                5name                            
         5 5name                0

SQL> select id,name,lead(name,2,0) over(order by id) from kkk;                                                                                                              
        ID NAME                 LEAD(NAME,2,0)OVER(ORDERBYID)    
---------- -------------------- -----------------------------    
         1 1name                3name                            
         2 2name                4name                            
         3 3name                5name                            
         4 4name                0                                
         5 5name                0 
SQL> select id,name,lead(name,1,'linjiqin') over(order by id) from kkk;                                 
                                                                                 
        ID NAME                 LEAD(NAME,1,'ALSDFJLASDJFSAF')                   
---------- -------------------- ------------------------------                   
         1 1name                2name                                            
         2 2name                3name                                            
         3 3name                4name                                            
         4 4name                5name                                            
         5 5name                linjiqin  

---------------------------------------------------------------------------------------

   

六、rollup()与cube():排列组合分组
1)、group by rollup(a, b, c):
首先会对(a、b、c)进行group by,
然后再对(a、b)进行group by,
其后再对(a)进行group by,
最后对全表进行汇总操作。

     

2)、group by cube(a, b, c):
则首先会对(a、b、c)进行group by,
然后依次是(a、b),(a、c),(a),(b、c),(b),(c),
最后对全表进行汇总操作。

   

1、生成演示数据:
Connected to Oracle Database 10g Enterprise Edition Release 10.2.0.1.0
Connected as ds_trade
 
SQL> conn system/oracle as sysdba
Connected to Oracle Database 10g Enterprise Edition Release 10.2.0.3.0
Connected as SYS
 
SQL> create table scott.t as select * from dba_indexes;
 
Table created
 
 
SQL> connect scott/oracle
Connected to Oracle Database 10g Enterprise Edition Release 10.2.0.3.0
Connected as scott
 
SQL>

    

2、普通group by体验
sql> select owner, index_type, status, count(*) from t where owner like 'SY%' group by owner, index_type, status;

3、group by rollup(A,B,C)
GROUP BY ROLLUP(A, B, C):
首先会对(A、B、C)进行GROUP BY,
然后再对(A、B)进行GROUP BY,
其后再对(A)进行GROUP BY,
最后对全表进行汇总操作。
sql> select owner, index_type, status, count(*) from t where owner like 'SY%' group by ROLLUP(owner, index_type, status);

4、group by cube(A,B,C)
GROUP BY CUBE(A, B, C):
则首先会对(A、B、C)进行GROUP BY,
然后依次是(A、B),(A、C),(A),(B、C),(B),(C),
最后对全表进行汇总操作。

sql> select owner, index_type, status, count(*) from t where owner like 'SY%' group by cube(owner, index_type, status);

  

七、max(),min(),sun()与avg():求移动的最值总和与平均值
问题:计算出各个地区连续3个月的通话费用的平均数(移动平均值)

 

SELECT AREA_CODE, 
BILL_MONTH,
LOCAL_FARE,
SUM(LOCAL_FARE) OVER(PARTITION BY AREA_CODE
ORDER BY TO_NUMBER(BILL_MONTH)
RANGE BETWEEN 1 PRECEDING AND 1 FOLLOWING) "3month_sum",
AVG(LOCAL_FARE) OVER(PARTITION BY AREA_CODE
ORDER BY TO_NUMBER(BILL_MONTH)
RANGE BETWEEN 1 PRECEDING AND 1 FOLLOWING) "3month_avg",
MAX(LOCAL_FARE) OVER(PARTITION BY AREA_CODE
ORDER BY TO_NUMBER(BILL_MONTH)
RANGE BETWEEN 1 PRECEDING AND 1 FOLLOWING) "3month_max",
MIN(LOCAL_FARE) OVER(PARTITION BY AREA_CODE
ORDER BY TO_NUMBER(BILL_MONTH)
RANGE BETWEEN 1 PRECEDING AND 1 FOLLOWING) "3month_min"
FROM (SELECT T.AREA_CODE, T.BILL_MONTH, SUM(T.LOCAL_FARE) LOCAL_FARE
FROM T
GROUP BY T.AREA_CODE, T.BILL_MONTH)

运行结果:

  

问题:求各地区按月份累加的通话费

SELECT AREA_CODE, 
BILL_MONTH,
LOCAL_FARE,
SUM(LOCAL_FARE) OVER(PARTITION BY AREA_CODE
ORDER BY BILL_MONTH ASC) "last_sum_value"
FROM (SELECT T.AREA_CODE, T.BILL_MONTH, SUM(T.LOCAL_FARE) LOCAL_FARE
FROM T
GROUP BY T.AREA_CODE, T.BILL_MONTH)
ORDER BY AREA_CODE, BILL_MONTH

运行结果:

 

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posted on 2012-04-06 15:43  Ruthless  阅读(10518)  评论(0编辑  收藏