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

浏览数：19 / 时间：2015年06月12日

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;