1.先显示一下ROLLUP的效果
sec@ora10g> 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.
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.再看一下CUBE的效果
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.
3.仔细观察一下,这儿两个的细微差别是什么?
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
sec@ora10g> 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.
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.再看一下CUBE的效果
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.
3.仔细观察一下,这儿两个的细微差别是什么?
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
来自 “ ITPUB博客 ” ,链接:http://blog.itpub.net/165278/viewspace-611086/,如需转载,请注明出处,否则将追究法律责任。
转载于:http://blog.itpub.net/165278/viewspace-611086/
ROLLUP与CUBE详解
本文详细对比了SQL中的ROLLUP和CUBE功能,并通过具体示例展示了它们在统计数据汇总时的区别。ROLLUP用于生成一系列子集的汇总,而CUBE则提供了所有可能的子集组合。
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