oracle 数据库开发面试题，当时笔试的时候一个没做出来，现附原题及答案

1、

ID
1
2
3
5
6
7
8
10
11
12
15

表名tt，用sql找出ID列中不连续的ID，例如其中没有的4：

--创建表及数据

CREATE TABLE tt(ID INTEGER);



INSERT INTO tt 

SELECT 1 FROM dual

UNION ALL

SELECT 2 FROM dual

UNION ALL

SELECT 3 FROM dual

UNION ALL

SELECT 5 FROM dual

UNION ALL

SELECT 6 FROM dual

UNION ALL

SELECT 7 FROM dual

UNION ALL

SELECT 8 FROM dual

UNION ALL

SELECT 10 FROM dual

UNION ALL

SELECT 11 FROM dual

UNION ALL

SELECT 12 FROM dual

UNION ALL

SELECT 15 FROM dual;

COMMIT;

 

--用到了connect by level 造数据

WITH IT AS

 (SELECT LEVEL ID FROM DUAL CONNECT BY LEVEL <= (SELECT MAX(ID) FROM TT))

SELECT A.ID

  FROM IT A

 WHERE NOT EXISTS (SELECT 1 FROM TT B WHERE A.ID = B.ID)

 

2、

将录入不规范的房间信息整理成规范格式

不规范表（多个房间用逗号分割）

ID	ROOM
1	101,102
2	201,202,203
3	301
.......

 

 

 

 

 

规范表

ID	ROOM
1	101
1	102
2	201
2	202
2	203
3	301
......

 

 

 

 

 

 

 

 

--单行单列转多行

--创建表及数据

create table ttt(id integer,room varchar2(200));



insert into ttt

select 1,'101,102' from dual

union all

select 2,'201,202,203' from dual

union all

select 3,'301' from dual;

commit;

SELECT DISTINCT ID,REGEXP_SUBSTR(room, '[^,]+', 1, LEVEL, 'i') AS STR  

  FROM ttt

CONNECT BY LEVEL <= LENGTH(room) - LENGTH(REGEXP_REPLACE(room, ',', ''))+1;