如何實現(xiàn)SQL查詢連續(xù)號碼段

這篇文章主要介紹“如何實現(xiàn)SQL查詢連續(xù)號碼段”，在日常操作中，相信很多人在如何實現(xiàn)SQL查詢連續(xù)號碼段問題上存在疑惑，小編查閱了各式資料，整理出簡單好用的操作方法，希望對大家解答”如何實現(xiàn)SQL查詢連續(xù)號碼段”的疑惑有所幫助！接下來，請跟著小編一起來學(xué)習(xí)吧！

我有一個表結(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

(第二個字段內(nèi)可能是連續(xù)的數(shù)據(jù)，可能存在斷點。)

怎樣能查詢出來這樣的結(jié)果,查詢出連續(xù)的記錄來。

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.

復(fù)制代碼 代碼如下:

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分析函數(shù)進行處理.. 樓上的方法確實挺好用就是覺得表掃描/表連接比較多, 可能數(shù)據(jù)量大了. 速度會比較慢, 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..

復(fù)制代碼 代碼如下:

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給了我這個答案, 頓時覺得耳目一新啊..就貼出來與大家一起共享了^_^.

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>

到此，關(guān)于“如何實現(xiàn)SQL查詢連續(xù)號碼段”的學(xué)習(xí)就結(jié)束了，希望能夠解決大家的疑惑。理論與實踐的搭配能更好的幫助大家學(xué)習(xí)，快去試試吧！若想繼續(xù)學(xué)習(xí)更多相關(guān)知識，請繼續(xù)關(guān)注億速云網(wǎng)站，小編會繼續(xù)努力為大家?guī)砀鄬嵱玫奈恼拢?/p>