静·谧——Last Winner
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这个好像这两个问题的逆问题
http://blog.itpub.net/post/7102/44843
http://blog.itpub.net/post/7102/48132


表A结构:
bill_type_id varchar2(1),
bill_start number,
bill_end number,
office_level varchar2(4)
数据如下:
A 0 999 1
A 0 199 2
A 300 499 2
A 700 799 2
sql目的是取出包含在level1级别里的,还没有录入level2级别的号段,各位大侠有好方法,请告知小弟.最好是一个sql语句实现,谢谢!

原文地址:http://www.itpub.net/480536.html


表结构以及测试数据

CREATE TABLE T8
(
A NUMBER(4),
B NUMBER(4),
C NUMBER(4),
Q VARCHAR2(1 BYTE)
);

Insert into T8
(A, B, C, Q)
Values
(555, 666, 2, 'A');
Insert into T8
(A, B, C, Q)
Values
(100, 199, 2, 'A');
Insert into T8
(A, B, C, Q)
Values
(0, 999, 1, 'A');
Insert into T8
(A, B, C, Q)
Values
(300, 499, 2, 'A');
COMMIT;

select S,E from
(
SELECT NVL2(LAG(A)OVER(PARTITION BY Q ORDER BY A),B+1,MIN(A)OVER(PARTITION BY Q)) S,
NVL(LEAD(A)OVER(PARTITION BY Q ORDER BY A)-1,MAX(B)OVER(PARTITION BY Q)) E
from t8 START WITH C=1 CONNECT BY C-1 = PRIOR C AND Q= PRIOR Q
)
where s<=e

代码:

SQL
> select * from t8;

A B C Q ---------- ---------- ---------- -
555 666 2 A
100 199 2 A
0 999 1 A
300 499 2 A
200 299 2 A

已用时间
: 00: 00: 00.00

Execution Plan
----------------------------------------------------------
0 SELECT STATEMENT Optimizer=CHOOSE
1 0 TABLE ACCESS
(FULL) OF 'T8'


SQL> select S,E from
2
(
3 SELECT NVL2(LAG(A)OVER(PARTITION BY Q ORDER BY A),B+1,MIN(A)OVER(PARTITION BY Q)) S,
4 NVL(LEAD(A)OVER(PARTITION BY Q ORDER BY A)-1,MAX(B)OVER(PARTITION BY Q)) E
5 from t8 START WITH C
=1 CONNECT BY C-1 = PRIOR C AND Q= PRIOR Q
6
)
7 where s<=e
8
/

S E ---------- ----------
0 99
500 554
667 999

已用时间
: 00: 00: 00.00

Execution Plan
----------------------------------------------------------
0 SELECT STATEMENT Optimizer=CHOOSE
1 0 VIEW
2 1 WINDOW
(SORT)
3 2 CONNECT BY (WITH FILTERING)
4 3 NESTED LOOPS
5 4 TABLE ACCESS
(FULL) OF 'T8'
6 4 TABLE ACCESS (BY USER ROWID) OF 'T8'
7 3 NESTED LOOPS
8 7 BUFFER
(SORT)
9 8 CONNECT BY PUMP
10 7 TABLE ACCESS
(FULL) OF 'T8'

.................


Q:level 1 级别的号段可能是分段的 如0-9999 20000-29999

A:关于分段的问题,你可以把数据增加一个分段标志
比如0~9999,20000~29999,,级别都为1,分段标志分别为1,2
假设级别2的数据有0~99, 300~1999, 21000~22999
那么前两个的分段标志就设为1,后一个就设为2
然后在上面的sql中,partition by后面加上这个分段标志就可以了
思路大致如此


select S,E from
(
SELECT NVL2(LAG(A)OVER(PARTITION BY Q, xx ORDER BY A),B+1,MIN(A)OVER(PARTITION BY Q, xx)) S,
NVL(LEAD(A)OVER(PARTITION BY Q, xx ORDER BY A)-1,MAX(B)OVER(PARTITION BY Q, xx)) E from
(
select t8.*, nvl(prior a,a) xx from t8 START WITH C=1 CONNECT BY C-1 = PRIOR C AND Q= PRIOR Q
and A>= prior a and b<= prior b
)
)
where s<=e

lastwinner 发表于:2006.01.16 23:59 ::分类: ( Oracle , ) ::阅读:(2749次) :: 评论 (1)
re: 关于号段选取的sql写法 [回复]

不知道where后面的那些层级语句有什么用,只要前半段语句不用也可以实现:

SQL> SELECT NVL2(LAG(A) OVER(PARTITION BY Q ORDER BY A),
2 B + 1,
3 MIN(A) OVER(PARTITION BY Q)) S,
4 NVL(LEAD(A) OVER(PARTITION BY Q ORDER BY A) - 1,
5 MAX(B) OVER(PARTITION BY Q)) E
6 from t8
7 ;

S E
---------- ----------
0 99
200 299
500 554
667 999

yaanzy 评论于: 2006.12.25 11:20

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