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詳細介紹分級匯總實現的3種方法的比較

2024-07-21 02:42:09
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分級匯總實現的3種方法比較

1.代碼示例:

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

select code 代碼 , substrb(' ',1,item_level*2-2)||b.reg_type 登記注冊類型, cnt 家數 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;

代碼 登記注冊類型 家數

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

100 內資企業

110 國有企業

120 集體企業

130 股份合作企業

140 聯營企業

141 國有聯營企業

142 集體聯營企業

143 國有與集體聯營企業

149 其他聯營企業

150 有限責任公司

151 國有獨資公司

159 其他有限責任公司

160 股份有限公司

170 私營企業

171 私營獨資企業

172 私營合伙企業

173 私營有限責任公司

174 私營股份有限公司

200 港、澳、臺商投資企業

210 合資經營企業(港或澳、臺資)

220 合作經營企業(港或澳、臺資)

230 港、澳、臺商獨資經營企業

240 港、澳、臺商投資股份有限公司

300 外商投資企業

310 中外合資經營企業

320 中外合作經營企業

330 外資企業

340 外商投資股份有限公司

----

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)

大家可以試試看。

2.代碼示例:

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

select code 代碼 , substrb(' ',1,item_level*2-2)||b.reg_type 登記注冊類型, cnt 家數 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

;

最終版14.89秒

代碼:------------------------------------------

select code 代碼 , substrb(' ',1,item_level*2-2)||b.reg_type 登記注冊類型, cnt 家數 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

;

在小一些的數據量上的執行情況:

3.代碼示例:

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

已連接。

SQL> set autot on

SQL> set timi on

SQL> select code 代碼 , substrb(' ',1,item_level*2-2)||b.reg_type 登記注冊類型, cnt 家數 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;

已選擇28行。

已用時間: 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 代碼 , substrb(' ',1,item_level*2-2)||b.reg_type 登記注冊類型, cnt 家數 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 ;

已選擇28行。

已用時間: 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 代碼 , substrb(' ',1,item_level*2-2)||b.reg_type 登記注冊類型, cnt 家數 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 ;

已選擇28行。

已用時間: 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>

大家可以發現,第3種的一致性取和物理讀都超過第2種,不過還是快一些。


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