分级汇总实现的3种方法比较
来源:互联网 发布:百兆网络用千兆交换机 编辑:程序博客网 时间:2024/06/05 15:10
分级汇总实现的3种方法比较
代码:--------------------------------------------------------------------------------
select code 代码 , substrb(' ',1,item_level*2-2)||b.reg_type 登记注册类型, cnt 家数 from
(
(select substr(z01_08,1,1)||'00' code ,count(*) cnt
from cj601
group by substr(z01_08,1,1))
union
(select substr(z01_08,1,2)||'0' code ,count(*) cnt
from cj601
group by substr(z01_08,1,2))
union
(select substr(z01_08,1,3) code ,count(*) cnt
from cj601
group by substr(z01_08,1,3))
)
c, djzclx b where c.code=b.reg_code;
代码 登记注册类型 家数
------ --------------------------------------- ---------
100 内资企业
110 国有企业
120 集体企业
130 股份合作企业
140 联营企业
141 国有联营企业
142 集体联营企业
143 国有与集体联营企业
149 其他联营企业
150 有限责任公司
151 国有独资公司
159 其他有限责任公司
160 股份有限公司
170 私营企业
171 私营独资企业
172 私营合伙企业
173 私营有限责任公司
174 私营股份有限公司
200 港、澳、台商投资企业
210 合资经营企业(港或澳、台资)
220 合作经营企业(港或澳、台资)
230 港、澳、台商独资经营企业
240 港、澳、台商投资股份有限公司
300 外商投资企业
310 中外合资经营企业
320 中外合作经营企业
330 外资企业
340 外商投资股份有限公司
----
lastwinner
type: substr(z01_08,1,1)||'00'
subtype : substr(z01_08,1,2)||'0'
sub-subtype : substr(z01_08,1,3)
select ..........
group by rollup(type, subtype, sub-subtype)
试试看
代码:
--------------------------------------------------------------------------------
select code 代码 , substrb(' ',1,item_level*2-2)||b.reg_type 登记注册类型, cnt 家数 from
(
select
case when code3 is not null then code3
when code2<>'0' then code2
else code1
end code,cnt
from (
select substr(z01_08,1,1)||'00' code1 , substr(z01_08,1,2)||'0' code2 , substr(z01_08,1,3) code3 ,count(*) cnt
from j601
group by rollup(substr(z01_08,1,1),substr(z01_08,1,2),substr(z01_08,1,3))
) where code2<>code3 or code3 is null and code1<>'00'
)
c, djzclx b where c.code=b.reg_code
order by 1
;
最终版14.89秒
代码:--------------------------------------------------------------------------------
select code 代码 , substrb(' ',1,item_level*2-2)||b.reg_type 登记注册类型, cnt 家数 from
(
select
case when code3 is not null then code3
when code2<>'0' then code2
else code1
end code,cnt
from (
select substr(z01_08,1,1)||'00' code1 , substr(z01_08,1,2)||'0' code2 , substr(z01_08,1,3) code3 ,sum(cnt) cnt
from (select substr(z01_08,1,3) z01_08,count(*) cnt from j601 group by substr(z01_08,1,3))
group by rollup(substr(z01_08,1,1),substr(z01_08,1,2),substr(z01_08,1,3))
) where code2<>code3 or code3 is null and code1<>'00'
)
c, djzclx b where c.code=b.reg_code
order by 1
;
在小一些的数据量上的执行情况
代码:--------------------------------------------------------------------------------
已连接。
SQL> set autot on
SQL> set timi on
SQL> select code 代码 , substrb(' ',1,item_level*2-2)||b.reg_type 登记注册类型, cnt 家数 from
2 (
3 (select substr(z01_08,1,1)||'00' code ,count(*) cnt
4 from cj601
5 group by substr(z01_08,1,1))
6 union
7 (select substr(z01_08,1,2)||'0' code ,count(*) cnt
8 from cj601
9 group by substr(z01_08,1,2))
10 union
11 (select substr(z01_08,1,3) code ,count(*) cnt
12 from cj601
13 group by substr(z01_08,1,3))
14 )
15 c, djzclx b where c.code=b.reg_code;
已选择28行。
已用时间: 00: 00: 01.03
Execution Plan
----------------------------------------------------------
0 SELECT STATEMENT Optimizer=CHOOSE
1 0 NESTED LOOPS
2 1 VIEW
3 2 SORT (UNIQUE)
4 3 UNION-ALL
5 4 SORT (GROUP BY)
6 5 TABLE ACCESS (FULL) OF 'CJ601'
7 4 SORT (GROUP BY)
8 7 TABLE ACCESS (FULL) OF 'CJ601'
9 4 SORT (GROUP BY)
10 9 TABLE ACCESS (FULL) OF 'CJ601'
11 1 TABLE ACCESS (BY INDEX ROWID) OF 'DJZCLX'
12 11 INDEX (UNIQUE SCAN) OF 'SYS_C002814' (UNIQUE)
Statistics
----------------------------------------------------------
199 recursive calls
0 db block gets
13854 consistent gets
2086 physical reads
0 redo size
1480 bytes sent via SQL*Net to client
514 bytes received via SQL*Net from client
3 SQL*Net roundtrips to/from client
8 sorts (memory)
0 sorts (disk)
28 rows processed
SQL> select code 代码 , substrb(' ',1,item_level*2-2)||b.reg_type 登记注册类型, cnt 家数 from
2 (
3 select
4 case when code3 is not null then code3
5 when code2<>'0' then code2
6 else code1
7 end code,cnt
8 from (
9 select substr(z01_08,1,1)||'00' code1 , substr(z01_08,1,2)||'0' code2 , substr(z01_08,1,3) code3 ,count(*) cnt
10 from cj601
11 group by rollup(substr(z01_08,1,1),substr(z01_08,1,2),substr(z01_08,1,3))
12 ) where code2<>code3 or code3 is null and code1<>'00'
13 )
14 c, djzclx b where c.code=b.reg_code
15 order by 1
16 ;
已选择28行。
已用时间: 00: 00: 00.07
Execution Plan
----------------------------------------------------------
0 SELECT STATEMENT Optimizer=CHOOSE
1 0 SORT (ORDER BY)
2 1 NESTED LOOPS
3 2 VIEW
4 3 FILTER
5 4 SORT (GROUP BY ROLLUP)
6 5 TABLE ACCESS (FULL) OF 'CJ601'
7 2 TABLE ACCESS (BY INDEX ROWID) OF 'DJZCLX'
8 7 INDEX (UNIQUE SCAN) OF 'SYS_C002814' (UNIQUE)
Statistics
----------------------------------------------------------
0 recursive calls
0 db block gets
4628 consistent gets
701 physical reads
0 redo size
1480 bytes sent via SQL*Net to client
514 bytes received via SQL*Net from client
3 SQL*Net roundtrips to/from client
2 sorts (memory)
0 sorts (disk)
28 rows processed
SQL> select code 代码 , substrb(' ',1,item_level*2-2)||b.reg_type 登记注册类型, cnt 家数 from
2 (
3 select
4 case when code3 is not null then code3
5 when code2<>'0' then code2
6 else code1
7 end code,cnt
8 from (
9 select substr(z01_08,1,1)||'00' code1 , substr(z01_08,1,2)||'0' code2 , substr(z01_08,1,3) code3 ,sum(cnt) cnt
10 from (select substr(z01_08,1,3) z01_08,count(*) cnt from cj601 group by substr(z01_08,1,3))
11 group by rollup(substr(z01_08,1,1),substr(z01_08,1,2),substr(z01_08,1,3))
12 ) where code2<>code3 or code3 is null and code1<>'00'
13 )
14 c, djzclx b where c.code=b.reg_code
15 order by 1
16 ;
已选择28行。
已用时间: 00: 00: 00.06
Execution Plan
----------------------------------------------------------
0 SELECT STATEMENT Optimizer=CHOOSE
1 0 SORT (ORDER BY)
2 1 NESTED LOOPS
3 2 VIEW
4 3 FILTER
5 4 SORT (GROUP BY ROLLUP)
6 5 VIEW
7 6 SORT (GROUP BY)
8 7 TABLE ACCESS (FULL) OF 'CJ601'
9 2 TABLE ACCESS (BY INDEX ROWID) OF 'DJZCLX'
10 9 INDEX (UNIQUE SCAN) OF 'SYS_C002814' (UNIQUE)
Statistics
----------------------------------------------------------
0 recursive calls
0 db block gets
4628 consistent gets
705 physical reads
0 redo size
1480 bytes sent via SQL*Net to client
514 bytes received via SQL*Net from client
3 SQL*Net roundtrips to/from client
3 sorts (memory)
0 sorts (disk)
28 rows processed
SQL>
第3种的一致性取和物理读都超过第2种,不过还是快一些
- 分级汇总实现的3种方法比较
- 分级查询的实现
- 分级控制打印的方法
- 树的实现 分级文件系统
- vc串口通信3种方法的实现和比较
- CRC16几种实现方法的比较
- grouping分级汇总用法
- SQL分级汇总
- 关于一个分级汇总
- sql分级汇总
- 实现下拉列表的分级显示
- html中动态分级表头的实现
- iOS开发 -- 延时执行某个方法的几种方式的比较和汇总
- JS实现跨域的几种方法汇总
- 多线程常用方法比较汇总
- java实现多线程的两种方法的比较
- android 图片旋转实现的两种方法的比较
- oracle数据库实现汇总报表的方法
- IIS+mysql+php配置详谈
- 翻译RFC3550-1.引言(introduction)
- 王垠:完全用Linux工作及其后续
- 如何备份注册表
- 内部网中的IP地址规划
- 分级汇总实现的3种方法比较
- 翻译RFC3550-2. RTP使用场景(RTP Use Scenarios)
- blog, blog.
- 密码破解速度全面披露
- PDA也玩无线:四款PDA无线上网卡导购
- 查询计算每个地区亏损前10名的语句的写法
- 糟糕的食物
- 用sqlldr从mysql导出一个表的数据到oracle
- 将汉化进行到底