oracle中CAST函数使用简介

CAST()函数可以进行数据类型的转换。

CAST()函数的参数有两部分，源值和目标数据类型，中间用AS关键字分隔。

以下例子均通过本人测试。

一、转换列或值

语法：cast( 列名/值 as 数据类型 )

用例：

1）、转换列

--将empno的类型（number）转换为varchar2类型。

select cast(empno as varchar2(10)) as empno from emp;

EMPNO
----------
7369
7499
7521
...

2)、转换值

--将字符串转换为整型。
SELECT CAST('123' AS int) as result from dual;

  RESULT
   ---

  123
返回值是整型值123。

--如果试图将一个代表小数的字符串转换为整型值，又会出现什么情况呢？
SELECT CAST('123.4' AS int) as result from dual;

RESULT
--------

  123

SELECT CAST('123.6' AS int) as result from dual;

RESULT
--------

  124
从上面可以看出，CAST()函数能执行四舍五入操作。

--截断小数

SELECT CAST('123.447654' AS decimal(5,2)) as result from dual;

RESULT
-----------
  123.45
decimal(5,2)表示值总位数为5，精确到小数点后2位。
SELECT CAST('123.4' AS decimal) as result from dual;
结果是一个整数值：
123
二、转换一个集合

语法：cast( multiset(查询语句) as 数据类型 )

1)转换成table

例子：

--学生成绩表

create table stu_score
(stu_no varchar2(50),--学号
  score  number--总分
  );
insert into stu_score values('201301',67);
insert into stu_score values('201302',63);
insert into stu_score values('201303',77);
insert into stu_score values('201304',68);
insert into stu_score values('201305',97);
insert into stu_score values('201306',62);
insert into stu_score values('201307',87);
commit;

------------------------------------------

select * from stu_score;

学号         分数

--------   ----------
201301       67
201302       63
201303       77
201304       68
201305       97
201306       62
201307       87

--奖学金表。

--奖学金表规定了名次，每个名次的人数和奖金。

create table scholarship
(
stu_rank   varchar(10),--名次
stu_num     int,--限定人数
money       number--奖金
);
insert into scholarship values('1',1,'1000');
insert into scholarship values('2',2,'500');
insert into scholarship values('3',3,'100');
commit；

-----------------------------------------------

select * from scholarship;

名次                                          人数     奖金
---------- --------------------------------------- ----------
1                                              1       1000
2                                              2        500
3                                              3        100

现在要根据成绩表的成绩降序排列，按奖学金表的名额确定排名和奖金。排名时不考虑相同成绩。
排名的结果应该如下：
学号          成绩        名次   奖金
201305        97          1        1000
201307        87           2        500
201303        77          2         500
201304        68          3         100
201301        67          3         100
201302        63          3         100

SELECT c.stu_no,c.score,b.stu_rank,b.money
   FROM (SELECT c.*,ROW_NUMBER() OVER(ORDER BY score DESC) rn FROM stu_score c) c
       ,(SELECT b.stu_rank,b.money,ROW_NUMBER() OVER(ORDER BY b.stu_rank) rn
          FROM scholarship b
             , TABLE( CAST( MULTISET( SELECT NULL
                                       FROM DUAL
                                    CONNECT BY LEVEL <= b.stu_num
                                    )
                             AS SYS.ODCIVARCHAR2LIST )
                            )
        ) b
WHERE c.rn=b.rn;

执行结果如下：

STU_NO                                                  SCORE      STU_RANK        MONEY
-------------------------------------------------- ----------         ----------          ----------
201305                                                     97                     1                1000
201307                                                     87                     2                 500
201303                                                     77                     2                 500
201304                                                     68                     3                 100
201301                                                     67                     3                 100
201302                                                     63                     3                 100

通过对比发现，确实达到了目的。

此外cast还能转化成collection，varray，此时都需要记过multiset集合函数一起使用。