[每日一题] OCP1z0-047 :2013-07-22 group by 子句

[每日一题] OCP1z0-047 :2013-07-22 group by 子句_第1张图片

[每日一题] OCP1z0-047 :2013-07-22 group by 子句_第2张图片

[每日一题] OCP1z0-047 :2013-07-22 group by 子句_第3张图片



 

这道题就是考where group by having的顺序。。。

答案A不正确:where应该放在group by前面

答案B不正确:having子句是用多行函数(sum,avg,max,min,count)等做为条件

答案C不正确:where应该放在group by前面

 

 

参考如下:(其实having 也可以放在group by前面)

 

SELECT   column, group_function

FROM      table

[WHERE   condition]

[GROUP BY group_by_expression]

[HAVING  group_condition]

[ORDER BY column];

 

答案:D


**********************************************************

一、group by分组注意:

drop table t;
drop table t1;
create table t(id number,dt date);
create table t1(id number,dt date);
insert into t values(1,sysdate);
insert into t1 values(1,sysdate+1);
insert into t1 values(1,sysdate+2);
commit;

--表中不正确
select t.id,max(t1.dt) mdt 
from t,t1 where t.id=t1.id;

--dual结合select 常量,始终返回一行,不管dual有多少行,所以正确
--在一般表里会出错,这个世界的规则有普通规则和特殊规则,如果都去遵守普遍规则,那么将会世界太平,特殊规则不用刻意去追求和遵循

with t as
(select 1 id,sysdate dt from dual),
t1 as
(select 1 id,sysdate+1 dt from dual union all
select 1,sysdate+2 from dual)
select t.id,max(t1.dt) mdt 
from t,t1 where t.id=t1.id;

--t1.id是多行,不行
with t as
(select 1 id,sysdate dt from dual),
t1 as
(select 1 id,sysdate+1 dt from dual
 UNION ALL 
select 1,sysdate+2 from dual)
select t1.id,max(t1.dt) mdt 
from t,t1 where t.id=t1.id;

--让tm来源于表,照样出错,和dual构造都有关系
with tm as
(select 1 id,sysdate dt from t),
t1 as
(select 1 id,sysdate+1 dt from dual union all
select 1,sysdate+2 from dual)
select t1.id,max(t1.dt) mdt 
from tm,t1 where tm.id=t1.id;

select dummy from dual;
--错误
select dummy from dual having count(*)=1;

--错误
with t as
(select dummy from dual)
select dummy from t having count(*)=1;

--用dummy就算取别名也出错,dummy列里不是常量??因为dummy允许多行
with t as
(select dummy x from dual)
select x from t having count(*)=1;
--正确,不用dummy,因为select 常量 from dual;不管dual有多少行,始终返回一行
with t as
(select 'X' x  from dual)
select x from t having count(*)=1;

select * from dual;
--给dual插入一条数据
insert into sys.dual values('Y');

----------------------------------------------------神奇的dual------------------------------------------------------------------
SQL> select dummy from dual;
 
DUMMY
-----
X
Y
 
SQL> select 'X' x from dual;
 
X
-
X
 
SQL> drop table m;
 
Table dropped
 
SQL> create table m(d varchar2(10));
 
Table created
 
SQL> insert into m values('a');
 
1 row inserted
 
SQL> insert into m values('b');
 
1 row inserted
 
SQL> select 'X' x from m;
 
X
-
X
X


二、having写在group by前后都一样


drop table t;
create table t(id number,name varchar2(10),sal number);
insert into t values(1,'a',2000);
insert into t values(1,'b',3000);
insert into t values(2,'c',1000);
insert into t values(2,'x',2000);
insert into t values(3,'d',5000);
insert into t values(4,'e',4000);
commit;
delete from t where id=1;
--下面的结果是一样的,但是最好用第2种,可读性强
select id,max(sal),count(*) from t having count(*)>1 group by id;
select id,max(sal),count(*) from t  group by id having count(*)>1;

俩个sql有 什么区别?

1.    select job,sum(sal) from emp t group by job having sum(sal) > 4200

2.   select job,sum(sal) from emp t having sum(sal) > 4200 group by job 

查询结果是一样的,具体这俩句有区别吗??

没有区别,oracle having可以放在前后,常规写法第1种



三、理解分组

http://www.itpub.net/thread-1042899-1-1.html


如何实现比较复杂的分组、小计与合计




--测试代码
create table t_dist
(
  TYPE_CD    NUMBER,
  BUYER_ID   VARCHAR2(50),
  ORDER_DT   DATE,
  SO_ID      VARCHAR2(50) not null,
  STOCK_ID   VARCHAR2(50) not null,
  UNIT_PRICE NUMBER,
  DISCOUNT   NUMBER,
  QTY        NUMBER
);


truncate table t_dist;
insert into t_dist values(1,'CN1001',to_date('2008-04-01','yyyy-mm-dd'),'S9001','29110311',50,10,8);
insert into t_dist values(1,'CN1001',to_date('2008-04-02','yyyy-mm-dd'),'S9002','29110312',60,20,2);
insert into t_dist values(1,'CN1001',to_date('2008-04-03','yyyy-mm-dd'),'S9003','29110313',70,15,3);
insert into t_dist values(2,'CN1001',to_date('2008-04-04','yyyy-mm-dd'),'S9004','29110312',60,15,5);
insert into t_dist values(2,'CN1001',to_date('2008-04-05','yyyy-mm-dd'),'S9005','29110311',70,10,6);
insert into t_dist values(3,'CN1001',to_date('2008-04-06','yyyy-mm-dd'),'S9006','29110313',55,20,4);
insert into t_dist values(3,'CN1001',to_date('2008-04-06','yyyy-mm-dd'),'S9007','29110311',40,10,3);
insert into t_dist values(3,'CN1001',to_date('2008-04-07','yyyy-mm-dd'),'S9008','29110312',50,50,5);
insert into t_dist values(3,'CN1001',to_date('2008-04-07','yyyy-mm-dd'),'S9009','29110313',80,10,2);
insert into t_dist values(1,'CN1001',to_date('2008-04-08','yyyy-mm-dd'),'S9010','29110311',65,10,1);
commit;


请问:如何实现如下结果,谢谢!
即计算按stock_id,type_cd,distount分组,计算每个产品的销售额(qty*unit_price)及销售百分比,并有小计


STOCK_ID        TYPE_CD        DISCOUNT     AVG_PRICE      SUM_TOT          PCT
-------------------------------------------------------------------------------------------------
29110311        1                        10              57.50             465              46.27%
29110311        2                        10              70.00             420              41.79%
29110311        3                        10              40.00             120              11.94%
小计                                                              55.83             1005            100.00%
29110312        1                        20              60.00             120              17.91%
29110312        2                        15              60.00             300              44.78%
29110312        3                        50              50.00             250              37.31%
小计                                                              56.67             670              100.00%
29110313        1                        15              70.00             210              35.59%
29110313        3                        10              80.00             160              27.12%
29110313        3                        20              55.00             220              37.29%
小计                                                              68.33             590              100.00%


哦, 我明白了PCT的意思了, 以后要把问题描述的清楚点![有没有奖励啊? ]


具体如下:






SQL> select case when grouping_id(type_cd,discount) = 3 then '小计' else stock_id end stock_id,
  2  type_cd,discount,avg(unit_price ) AVG_PRICE,
  3  sum(qty*unit_price) SUM_TOT,
  4  RATIO_TO_REPORT(sum(qty*unit_price )) over(partition by stock_id)*2 PCT
  5   from t_dist
  6  group by stock_id,rollup((type_cd,discount));


STOCK_ID                                        TYPE_CD   DISCOUNT  AVG_PRICE    SUM_TOT        PCT
-------------------------------------------------- ---------- ---------- ---------- ---------- ----------
29110311                                              1         10       57.5        465 0.46268656
29110311                                              2         10         70        420 0.41791044
29110311                                              3         10         40        120 0.11940298
小计                                                                        56.25       1005          1
29110312                                              1         20         60        120 0.17910447
29110312                                              2         15         60        300 0.44776119
29110312                                              3         50         50        250 0.37313432
小计                                                                   56.6666666        670          1
29110313                                              1         15         70        210 0.35593220
29110313                                              3         10         80        160 0.27118644
29110313                                              3         20         55        220 0.37288135
小计                                                                   68.3333333        590          1


12 rows selected




QUOTE:
--------------------------------------------------------------------------------
原帖由 dingjun123 于 2011-2-23 10:13 发表 


理解分组的概念
sum(sum(........之后相当于什么样的分组,后面又来个sum(.............


当然报错了 
--------------------------------------------------------------------------------




还是不理解 ,
暂不看sum,这里的 分子。分母是一样的啊。只是sum了分母,还是sum了分子;
相当于 sum(sum(a))/sum(a)  与 sum(a)/sum(sum(a)) ,后面的行的通,前面的怎么可能行不通啊




理解这个
with t as
(select mod(level,2) id from dual connect by level<10 )
select id,sum(id) from t
group by id;


with t as
(select mod(level,2) id from dual connect by level<10 )
select-- id,
sum(sum(id)) from t
group by id;


第2句为什么不能有id,因为sum(sum,外面的sum相当于全量分组了,相当于


with t as
(select mod(level,2) id from dual connect by level<10 )
select sum(x) from (
select id,
sum(id) x from t
group by id
);


那么当然不能有非汇总列在select里显示啊


一针见血啊。高~,很耐心的指导啊


还有个疑问
select 
sum(sum(QTY )) over (partition by stock_Id) PCT1 ,
sum(sum(QTY )) over (partition by stock_Id,grouping(type_cd)) PCT2
  from  t_dist
group by rollup(stock_id,(type_cd,DISCOUNT))


PCT1,PCT2有何区别?我实在是想不懂了。


等价于这个
select sum(x)  over (partition by stock_Id) PCT1,
sum(x) over(partition by stock_Id,gp) PCT2
from (
select 
sum(QTY ) x,stock_Id,grouping(type_cd) gp
  from  t_dist
group by rollup(stock_id,(type_cd,DISCOUNT))
);


看红色部分就知道第3个加了grouping的值,那么分区(分组)是不同的,所以两个结果不同


 

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