昨天在itpub看到這個(gè)帖子, 問(wèn)題覺(jué)得有意思,, 就仔細(xì)想了想. 也給出了一種解決辦法..:-)
問(wèn)題求助,請(qǐng)高手指點(diǎn)..
我有一個(gè)表結(jié)構(gòu)，
fphm,kshm
2014,00000001
2014,00000002
2014,00000003
2014,00000004
2014,00000005
2014,00000007
2014,00000008
2014,00000009
2013,00000120
2013,00000121
2013,00000122
2013,00000124
2013,00000125
(第二個(gè)字段內(nèi)可能是連續(xù)的數(shù)據(jù)，可能存在斷點(diǎn)。)
怎樣能查詢(xún)出來(lái)這樣的結(jié)果,查詢(xún)出連續(xù)的記錄來(lái)。
就像下面的這樣？
2014,00000001,00000005
2014,00000009,00000007
2013,00000120,00000122
2013,00000124,00000125
方法一: 引用自hmxxyy.
ID SEQ
---------- ----------
1 1
1 4
1 5
1 8
2 1
2 2
2 9
select res1.id, res2.seq str, res1.seq end
from (
select rownum rn, c.*
from (
select *
from gap a
where not exists (
select null from gap b where b.id = a.id and a.seq = b.seq - 1
)
order by id, seq
) c
) res1, (
select rownum rn, d.*
from (
select *
from gap a
where not exists (
select null from gap b where b.id = a.id and a.seq = b.seq + 1
)
order by id, seq
) d
) res2
where res1.id = res2.id
and res1.rn = res2.rn
/
ID STR END
--------- ---------- ----------
1 1 1
1 4 5
1 8 8
2 1 2
2 9 9
方法二: 使用lag/lead分析函數(shù)進(jìn)行處理.. 樓上的方法確實(shí)挺好用就是覺(jué)得表掃描/表連接比較多, 可能數(shù)據(jù)量大了. 速度會(huì)比較慢, 當(dāng)然我的這種方法由于使用分析函數(shù)使用的比較頻繁.所以排序量可能比上一種要多..
FPHM KSHM
---------- ----------------
2014 00000001
2014 00000002
2014 00000003
2014 00000004
2014 00000005
2014 00000007
2014 00000008
2014 00000009
2013 00000120
2013 00000121
2013 00000122
FPHM KSHM
---------- ----------------
2013 00000124
2013 00000125
13 rows selected.
SQL> set echo on
SQL> @bbb.sql
SQL> select fphm,lpad(kshm,8,'0') start_kshm,lpad(prev_prev_kshm,8,'0') end_kshm
2 from (
3 select fphm,kshm,next_kshm,prev_kshm,
4 lag(kshm,1,null) over (partition by fphm order by kshm )next_next_kshm,
5 lead(kshm,1,null) over (partition by fphm order by kshm ) prev_prev_kshm
6 from (
7 select *
8 from (
9 select fphm,kshm,
10 lead(kshm,1,null) over (partition by fphm order by kshm) next_kshm,
11 lag(kshm,1,null) over (partition by fphm order by kshm) prev_kshm
12 from t
13 )
14 where ( next_kshm - kshm <> 1 or kshm - prev_kshm <> 1 )
15 or ( next_kshm is null or prev_kshm is null )
16 )
17 )
18 where next_kshm - kshm = 1
19 /
FPHM START_KSHM END_KSHM
---------- ---------------- ----------------
2013 00000120 00000122
2013 00000124 00000125
2014 00000001 00000005
2014 00000007 00000009
SQL> spool off
方法三: 今天早上wildflower給了我這個(gè)答案, 頓時(shí)覺(jué)得耳目一新啊..就貼出來(lái)與大家一起共享了^_^.
SQL> spool aaa.log
SQL> set echo on
SQL> select * from t;
no rows selected
SQL> select * from t;
FPHM KSHM
---------- ----------
2014 1
2014 2
2014 3
2014 4
2014 5
2014 7
2014 8
2014 9
2013 120
2013 121
2013 122
FPHM KSHM
---------- ----------
2013 124
2013 125
13 rows selected.
SQL> @bbb.sql
SQL> select b.fphm,min(b.kshm),max(b.kshm)
2 from (
3 select a.*,to_number(a.kshm-rownum) cc
4 from (
5 select * from t order by fphm,kshm
6 ) a
7 ) b
8 group by b.fphm,b.cc
9 /
FPHM MIN(B.KSHM) MAX(B.KSHM)
---------- ----------- -----------
2013 120 122
2013 124 125
2014 1 5
2014 7 9
SQL>
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