char的长度是固定的，而VARCHAR2的长度是可以变化的.
比如，存储字符串“abc"，对于CHAR (20)，表示你存储的字符将占20个字节(包括17个空字符)，
而同样的VARCHAR2 (20)则只占用3个字节的长度，20只是最大值，当你存储的字符小于20时，按实际长度存储。
char转化成varchar2会在字符串后补充空格 . 所以如果将char类型修改为varchar2类型需要使用RTRIM函数删除字段末尾补充的空格。
以下引用 自 MOS:
SOLUTION
EXAMPLE:
Note that the job column is currently a char column with a length of 11.
SQL> desc empcon
Name Null? Type
----------------------------------------- -------- --------------------------
EMPNO NUMBER(4)
ENAME VARCHAR2(10)
JOB CHAR(11 CHAR)
MGR NUMBER(4)
HIREDATE DATE
SAL NUMBER(7,2)
COMM NUMBER(7,2)
DEPTNO NUMBER(2)
SQL> alter table empcon modify job varchar2 (11);
(This alters the column to varchar2 but the length must be the same as the char column.)
Table altered.
SQL> desc empcon
Name Null? Type
----------------------------------------- -------- --------------------------
EMPNO NUMBER(4)
ENAME VARCHAR2(10)
JOB VARCHAR2(11)
MGR NUMBER(4)
HIREDATE DATE
SAL NUMBER(7,2)
COMM NUMBER(7,2)
DEPTNO NUMBER(2)
SQL> update empcon e
2 set e.job = RTRIM(e.job);
(This removes the blanks used to pad the char column)
14 rows updated.
SOLUTION
=======================
To implement the solution, please execute the following steps:
1. Modify the column to be a varchar2.
2. Use the update statement with the rtrim function to remove the blanks on the varchar2 column.
Please note that trying to insert into a varchar2 column of lesser length than the char will result in the following error:
ORA-12899: value too large for column "SCOTT"."EMPCON"."JOB" (actual: 11, maximum: 9)
