一个关于oracle中的函数问题

Question

SQL>  select ceil(3.1) from dual;CEIL(3.1)
---------
        4select decode(sign(n-3),-1,'n<3','n>=3') from tbname;

Answer 1

SQL>  select ceil(3.1) from dual;CEIL(3.1)
---------
        4select decode(sign(n-3),-1,'n<3','n>=3') from tbname;

Answer 2

SQL>  select ceil(3.1) from dual;CEIL(3.1)
---------
        4select decode(sign(n-3),-1,'n<3','n>=3') from tbname;

Answer 3

round(3.1)=3 round(3.5)=4 四舍五入 
trunc(3.1)=3  取整decode(sign(var-3),1，var比3大时取的值，-1，var比3大时取的值）

Answer 4

取整有向上，向下向下：
SQL> select trunc(3.1) from dual;
TRUNC(3.1)
----------
         3
SQL> select floor(3.1) from dual;
FLOOR(3.1)
----------
         3
SQL> select round(3.1) from dual;
ROUND(3.1)
----------
         3
向上：
SQL> select ceil(3.1) from dual;
 CEIL(3.1)
----------
         4