将数据分级汇总实现的3种方法比较

2009-12-14 18:51| 发布者: admin| 查看: 33| 评论: 0|原作者: 慕容紫英

★代码:-----------------......

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

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将数据分级汇总实现的3种方法比较

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