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详细介绍分级汇总实现的3种方法的比较

发布时间:2019-09-13 19:17:31 编辑:笔名

1.代码示例:

--------------------------------------------------------

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)

大家可以试试看。

2.代码示例:

-----------------------------------------------------

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 code2code3 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 code2code3 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:

3.代码示例:

--------------------------------------

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 code2code3 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 code2code3 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>

大家可以发现,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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