http://blog.csdn.net/hyde114/article/details/5135271
在处理数据时发现一个问题，Oracle对于小于1的小数，小数点前面的0是不显示的。
SQL> create table t1_number (num number);
Table created
SQL> insert into t1_number values(0.3268);
1 row inserted
SQL> insert into t1_number values(0.57965686);
1 row inserted
SQL> insert into t1_number values(52345234.5686);
1 row inserted
SQL> insert into t1_number values(4.552686);
1 row inserted
SQL> commit;
Commit complete
Table created
SQL> insert into t1_number values(0.3268);
1 row inserted
SQL> insert into t1_number values(0.57965686);
1 row inserted
SQL> insert into t1_number values(52345234.5686);
1 row inserted
SQL> insert into t1_number values(4.552686);
1 row inserted
SQL> commit;
Commit complete
SQL> column num format 99999999999.9999999999999
SQL> select * from t1_number;
SQL> select * from t1_number;
NUM
--------------------------
.3268000000000
.5796568600000
52345234.5686000000000
4.5526860000000
--------------------------
.3268000000000
.5796568600000
52345234.5686000000000
4.5526860000000
使用PL/SQL DEV工具查看时是有前面的0的
通过column num format 000000000000.99999999999设置也可以显示前面的0
但是这些都只是显示的结果，而不是数据库保存结果
SQL> column num format 00000000000.9999999999999
SQL> select * from t1_number;
SQL> select * from t1_number;
NUM
--------------------------
00000000000.3268000000000
00000000000.5796568600000
00052345234.5686000000000
00000000004.5526860000000
00000000005.2920000000000
--------------------------
00000000000.3268000000000
00000000000.5796568600000
00052345234.5686000000000
00000000004.5526860000000
00000000005.2920000000000
SQL> select substr(num,1,1) from t1_number;
SU
--
.
.
5
4
5
--
.
.
5
4
5
可以看到，无论前面出现多少位，通过substr函数获取的第一位任然是'.'
无论怎么转换，只要最终输出结果是数值型的，小数点前的0必然被丢弃，所以只有将数值转换为字符型，大致有两种方法：
1、直接使用to_char函数
首先实验一下将小数点前和小数点后都加上99999999的格式：
SQL> select to_char(num,'999999999.999999999') from t1_number;
TO_CHAR(NUM,'9999999
--------------------
.326800000
.579656860
52345234.568600000
4.552686000
5.292000000
--------------------
.326800000
.579656860
52345234.568600000
4.552686000
5.292000000
发现小数点前还是没有0，而小数点后的0多出来了
再实验一下小数点前和小数点后都加上0000000000格式：
SQL> select to_char(num,'00000000000.000000000')
from t1_number;
from t1_number;
TO_CHAR(NUM,'000000000
----------------------
00000000000.326800000
00000000000.579656860
00052345234.568600000
00000000004.552686000
00000000005.292000000
----------------------
00000000000.326800000
00000000000.579656860
00052345234.568600000
00000000004.552686000
00000000005.292000000
发现小数点前是有0了，但是两边的0都多出来了，使用FM来去掉多于的0：
SQL> select
to_char(num,'fm00000000000.000000000') from t1_number;
to_char(num,'fm00000000000.000000000') from t1_number;
TO_CHAR(NUM,'FM0000000
----------------------
00000000000.326800000
00000000000.579656860
00052345234.568600000
00000000004.552686000
00000000005.292000000
----------------------
00000000000.326800000
00000000000.579656860
00052345234.568600000
00000000004.552686000
00000000005.292000000
没有效果，说明FM只能去掉用9表示的格式中产生的0：
SQL> select
to_char(num,'fm999999999.999999999') from t1_number;
to_char(num,'fm999999999.999999999') from t1_number;
TO_CHAR(NUM,'FM99999
--------------------
.3268
.57965686
52345234.5686
4.552686
5.292
--------------------
.3268
.57965686
52345234.5686
4.552686
5.292
但是这样还是没有0，最后发现可以将小数点前的第一位置为0即可(注意9的个数要大于数值的位数)：
SQL> select
to_char(num,'fm999999990.999999999') from t1_number;
to_char(num,'fm999999990.999999999') from t1_number;
TO_CHAR(NUM,'FM99999
--------------------
0.3268
0.57965686
52345234.5686
4.552686
5.292
--------------------
0.3268
0.57965686
52345234.5686
4.552686
5.292
检查一下第一位，没有问题：
SQL> select
substr(to_char(num,'fm999999990.999999999'),1,1) from t1_number;
substr(to_char(num,'fm999999990.999999999'),1,1) from t1_number;
SU
--
0
0
5
4
5
--
0
0
5
4
5
2、使用decode函数
既然小于1的小数首位必然是'.'，那就判断首位是否为'.'，是则在前面加上'0'即可
SQL> select
decode(substr(num,1,1),'.','0'||num,num) from t1_number
decode(substr(num,1,1),'.','0'||num,num) from t1_number
DECODE(SUBSTR(NUM,1,1),'.','0'||NUM,NUM)
-----------------------------------------
0.3268
0.57965686
52345234.5686
4.552686
5.292
同样检查一下第一位，没有问题
SQL> select
substr(decode(substr(num,1,1),'.','0'||num,num),1,1) fro
m t1_number;
substr(decode(substr(num,1,1),'.','0'||num,num),1,1) fro
m t1_number;
SU
--
0
0
5
4
5
--
0
0
5
4
5
也可以写个函数来判断一下，不过可以用decode直接出来，感觉也没什么必要，暂时就这些吧。