如何实现SQL查询连续号码段

这篇文章主要介绍"如何实现SQL查询连续号码段"，在日常操作中，相信很多人在如何实现SQL查询连续号码段问题上存在疑惑，小编查阅了各式资料，整理出简单好用的操作方法，希望对大家解答"如何实现SQL查询连续号码段"的疑惑有所帮助！接下来，请跟着小编一起来学习吧！

我有一个表结构，
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

(第二个字段内可能是连续的数据，可能存在断点。)

怎样能查询出来这样的结果,查询出连续的记录来。

就像下面的这样？

2014,00000001,00000005
2014,00000009,00000007
2013,00000120,00000122
2013,00000124,00000125

方法一: 引用自hmxxyy.

复制代码 代码如下:

SQL> select * from gap;

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分析函数进行处理.. 楼上的方法确实挺好用就是觉得表扫描/表连接比较多, 可能数据量大了. 速度会比较慢, 当然我的这种方法由于使用分析函数使用的比较频繁.所以排序量可能比上一种要多..

复制代码 代码如下:

SQL> select fphm,lpad(kshm,8,'0') kshm
2 from t
3 /

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给了我这个答案, 顿时觉得耳目一新啊..就贴出来与大家一起共享了^_^.

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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