Oracle的CUBE与ROLLUP功能很相似，也是在数据统计分析领域的一把好手。
关于ROLLUP的查询统计功能请参考文章《Oracle分组函数之ROLLUP魅力》(）。
1.先看一下ROLLUP的数据统计效果
1）创建测试表group_test
SECOOLER@ora11g> create table group_test (group_id int, job varchar2(10), name varchar2(10), salary int);
Table created.
2）初始化数据
insert into group_test values (10,'Coding', 'Bruce',1000);
insert into group_test values (10,'Programmer','Clair',1000);
insert into group_test values (10,'Architect', 'Gideon',1000);
insert into group_test values (10,'Director', 'Hill',1000);
insert into group_test values (20,'Coding', 'Jason',2000);
insert into group_test values (20,'Programmer','Joey',2000);
insert into group_test values (20,'Architect', 'Martin',2000);
insert into group_test values (20,'Director', 'Michael',2000);
insert into group_test values (30,'Coding', 'Rebecca',3000);
insert into group_test values (30,'Programmer','Rex',3000);
insert into group_test values (30,'Architect', 'Richard',3000);
insert into group_test values (30,'Director', 'Sabrina',3000);
insert into group_test values (40,'Coding', 'Samuel',4000);
insert into group_test values (40,'Programmer','Susy',4000);
insert into group_test values (40,'Architect', 'Tina',4000);
insert into group_test values (40,'Director', 'Wendy',4000);
commit;
3）初始化之后的数据情况如下：
SECOOLER@ora11g> set pages 100
SECOOLER@ora11g> select * from group_test;
GROUP_ID JOB NAME SALARY
---------- ---------- ---------- ----------
10 Coding Bruce 1000
10 Programmer Clair 1000
10 Architect Gideon 1000
10 Director Hill 1000
20 Coding Jason 2000
20 Programmer Joey 2000
20 Architect Martin 2000
20 Director Michael 2000
30 Coding Rebecca 3000
30 Programmer Rex 3000
30 Architect Richard 3000
30 Director Sabrina 3000
40 Coding Samuel 4000
40 Programmer Susy 4000
40 Architect Tina 4000
40 Director Wendy 4000
16 rows selected.
4）ROLLUP的数据统计效果
sec@ora10g> select group_id,job,grouping(GROUP_ID),grouping(JOB),sum(salary) from group_test group by rollup(group_id, job);
GROUP_ID JOB GROUPING(GROUP_ID) GROUPING(JOB) SUM(SALARY)
---------- ---------- ------------------ ------------- -----------
10 Coding 0 0 1000
10 Director 0 0 1000
10 Architect 0 0 1000
10 Programmer 0 0 1000
10 0 1 4000
20 Coding 0 0 2000
20 Director 0 0 2000
20 Architect 0 0 2000
20 Programmer 0 0 2000
20 0 1 8000
30 Coding 0 0 3000
30 Director 0 0 3000
30 Architect 0 0 3000
30 Programmer 0 0 3000
30 0 1 12000
40 Coding 0 0 4000
40 Director 0 0 4000
40 Architect 0 0 4000
40 Programmer 0 0 4000
40 0 1 16000
1 1 40000
21 rows selected.
2.进一步体验CUBE的魅力
sec@ora10g> select group_id,job,grouping(GROUP_ID),grouping(JOB),sum(salary) from group_test group by cube(group_id, job) order by 1;
GROUP_ID JOB GROUPING(GROUP_ID) GROUPING(JOB) SUM(SALARY)
---------- ---------- ------------------ ------------- -----------
10 Architect 0 0 1000
10 Coding 0 0 1000
10 Director 0 0 1000
10 Programmer 0 0 1000
10 0 1 4000
20 Architect 0 0 2000
20 Coding 0 0 2000
20 Director 0 0 2000
20 Programmer 0 0 2000
20 0 1 8000
30 Architect 0 0 3000
30 Coding 0 0 3000
30 Director 0 0 3000
30 Programmer 0 0 3000
30 0 1 12000
40 Architect 0 0 4000
40 Coding 0 0 4000
40 Director 0 0 4000
40 Programmer 0 0 4000
40 0 1 16000
Architect 1 0 10000
Coding 1 0 10000
Director 1 0 10000
Programmer 1 0 10000
1 1 40000
25 rows selected.
解释如上结果中GROUPING函数返回值“0”和“1”的含义。
如果显示“1”表示CUBE函数对应的列（例如JOB字段）是由于CUBE函数所产生的空值对应的信息，即对此列进行汇总计算后的结果。
如果显示“0”表示此行对应的这列参未与ROLLUP函数分组汇总活动。
如果还是没有理解清楚，请参见Oracle官方文档中的描述内容：“Using a single column as its argument,GROUPINGreturns 1 when it encounters aNULLvalue created by aROLLUPorCUBEoperation. That is, if theNULLindicates the row is a subtotal,GROUPINGreturns a 1. Any other type of value, including a storedNULL, returns a 0.”
3.仔细观察一下，CUBE与ROLLUP之间的细微差别
rollup(a,b) 统计列包含：(a,b)、(a)、()
rollup(a,b,c) 统计列包含：(a,b,c)、(a,b)、(a)、()
……以此类推ing……
cube(a,b) 统计列包含：(a,b)、(a)、(b)、()
cube(a,b,c) 统计列包含：(a,b,c)、(a,b)、(a,c)、(b,c)、(a)、(b)、(c)、()
……以此类推ing……
So，上面例子中CUBE的结果比ROLLUP多了下面关于第一列GROUP_ID的统计信息：
Architect 1 0 10000
Coding 1 0 10000
Director 1 0 10000
4.小结
CUBE在ROLLUP的基础上进一步从各种维度上给出细化的统计汇总结果。
CUBE与GROUP BY的关系可以参考Oracle官方文档中的例子，链接如下：，链接如下：《CUBE Extension to GROUP BY》http://docs.oracle.com/cd/E11882_01/server.112/e25554/aggreg.htm#DWHSG8614