在ORACLE数据库中，NUMBER(P,S)是最常见的数字类型，可以存放数据范围为10^-130~10^126（不包含此值)，需要1~22字节(BYTE)不等的存储空间。P 是Precison的英文缩写，即精度缩写，表示有效数字的位数，最多不能超过38个有效数字。S是Scale的英文缩写，表示从小数点到最低有效数字的位数，它为负数时，表示从最大有效数字到小数点的位数。有时候，我们在创建表的时候，NUMBER往往没有指定P，S的值，那么默认情况下，NUMBER的P、S的值分别是多少呢？相信这个问题能问倒一大片DBA。 在之前，我遇到了一个问题，总结整理在“ORACLE NUMBER类型Scale为0引发的问题”这篇博客当中，当时武断的判断“如果不指定p和s，NUMBER类型，它的默认精度值为38, 默认的scale值为0”，因为当时参考了官方文档https://docs.oracle.com/cd/B28359_01/server.111/b28318/datatype.htm#CNCPT1832
当然文档没有错误，文档应该是指在定义字段数据类型为NUMBER时，指定了NUMBER类型的P值，但是没有指定S的值，那么Scale默认就是0，如下测试所示，当时应该是我自己没有完全理解文档意思，当然文档也有误导的嫌疑。
SQL> drop table test;
Table dropped.SQL> create table test(id number(38));
Table created.SQL> insert into test
2 select 123 from dual union all
3 select 123.123 from dual;
2 rows created.SQL> commit;
Commit complete.SQL> select * from test;
ID
---------- 123123
SQL>
当在指定字段类型为NUMBER时，如果P和S都不指定，那么P和S又是什么值呢？今天特意实验验证了一下，具体实验过程如下：
SQL> drop table test;
Table dropped.SQL> create table test(id number, id1 number(38,4));
Table created.SQL> insert into test
2 select 12, 12 from dual union all
3 select 12.123456789, 12.123456789 from dual;
2 rows created.SQL> commit;
Commit complete.SQL> col id for 999999999999.999999999999999999999999999999999999;
SQL> col id1 for 99999999999.9999999999999999999999999999999999999;
SQL> select * from test;
ID ID1
-------------------------------------------------- -------------------------------------------------- 12.000000000000000000000000000000000000 12.000000000000000000000000000000000000012.123456789000000000000000000000000000 12.1235000000000000000000000000000000000
SQL>
如上所示，当我插入上面两条记录后，发现如果不指定p和s，NUMBER类型，此时的Scale至少是9，我们继续测试，插入下面数据
SQL> insert into test
2 select 12.123456789123456789123456789123456,3 12.123456789123456789123456789123456
4 from dual;1 row created.SQL> commit;
Commit complete.SQL> select * from test;
ID ID1
-------------------------------------------------- -------------------------------------------------- 12.000000000000000000000000000000000000 12.000000000000000000000000000000000000012.123456789000000000000000000000000000 12.1235000000000000000000000000000000000
12.123456789123456789123456789123456000 12.1235000000000000000000000000000000000
如下所示，此时可以看到Scale的值33了，那么Scale的值是否可以继续变大呢？
SQL> insert into test
2 select 12.123456789123456789123456789123456789123,3 12.123456789123456789123456789123456789123
4 from dual;1 row created.SQL> commit;
Commit complete.SQL> select * from test;
ID ID1
-------------------------------------------------- -------------------------------------------------- 12.000000000000000000000000000000000000 12.000000000000000000000000000000000000012.123456789000000000000000000000000000 12.1235000000000000000000000000000000000
12.123456789123456789123456789123456000 12.1235000000000000000000000000000000000
12.123456789123456789123456789123456789 12.1235000000000000000000000000000000000
如下截图所示，插入的记录为12.123456789123456789123456789123456789123，但是显示的值为12.123456789123456789123456789123456789，总共为38位，由于格式化列的缘故，可能导致部分小数位没有显示，
我们继续测试，调整格式化列，我们发现值变为了12.12345678912345678912345678912345678912，总共40位了，Scale的值为38了。这个是为什么呢？不是数字精度为38，意味着最多是38位吗？
SQL> col id for 999999999999.99999999999999999999999999999999999999999
SQL> col id1 for 99999999999.999999999999999999999999999999999999999999
SQL> select * from test;
ID ID1
------------------------------------------------------- ------------------------------------------------------- 12.00000000000000000000000000000000000000000 12.00000000000000000000000000000000000000000012.12345678900000000000000000000000000000000 12.123500000000000000000000000000000000000000
12.12345678912345678912345678912345600000000 12.123500000000000000000000000000000000000000
12.12345678912345678912345678912345678912000 12.123500000000000000000000000000000000000000
继续其它测试，我们发现Sacle的值会随着小数点前面数字的变化而变化，如下所示：
SQL> insert into test
2 select 123456789.123456789123456789123456789123456,3 123456789.123456789123456789123456789123456
4 from dual;1 row created.SQL> commit;
Commit completeSQL> insert into test
2 select 123456789123.123456789123456789123456789123456,3 123456789123.123456789123456789123456789123456
4 from dual;1 row created.SQL> commit;
Commit complete.SQL> select * from test;
ID ID1
------------------------------------------------------- ------------------------------------------------------- 12.00000000000000000000000000000000000000000 12.00000000000000000000000000000000000000000012.12345678900000000000000000000000000000000 12.123500000000000000000000000000000000000000
12.12345678912345678912345678912345600000000 12.123500000000000000000000000000000000000000
12.12345678912345678912345678912345678912000 12.123500000000000000000000000000000000000000
123456789.12345678912345678912345678912300000000000 123456789.123500000000000000000000000000000000000000
123456789123.12345678912345678912345678910000000000000 #######################################################
6 rows selected.
从上面测试可以看出，Scale的值是变化的，跟数据值有关系，目前看来，小数点前的数字位数和小数点后的数字位数相加为40（有时候又是39），为了测试是否这个规律，我特意用下面案例测试一下
SQL> create table test2(id number);
Table created.SQL> insert into test2
2 select 0.123456789123456789123456789123456789123456789 from dual;
1 row created.SQL> commit;
Commit complete.SQL> col id for 9999999999.9999999999999999999999999999999999999999999999;
SQL> select * from test2;
ID
---------------------------------------------------------- .1234567891234567891234567891234567891235000000SQL> insert into test2
2 select 123456789.123456789123456789123456789123456789 from dual;
1 row created.SQL> commit;
Commit complete.SQL> select * from test2;
ID
---------------------------------------------------------- .1234567891234567891234567891234567891235000000123456789.1234567891234567891234567891230000000000000000
SQL> insert into test2
2 select 123456789123.123456789123456789123456789123456789 from dual;
1 row created.SQL> commit;
Commit complete.SQL> select * from test2;
ID
---------------------------------------------------------- .1234567891234567891234567891234567891235000000123456789.1234567891234567891234567891230000000000000000
##########################################################
SQL> col id for 9999999999999.9999999999999999999999999999999999999999999999;
SQL> select * from test2;
ID
------------------------------------------------------------- .1234567891234567891234567891234567891235000000123456789.1234567891234567891234567891230000000000000000
123456789123.1234567891234567891234567891000000000000000000
SQL> insert into test2
2 select 12345678912345.12345678912345678912345678912345 from dual;
1 row created.SQL> commit;
Commit complete.SQL> select * from test2;
ID
------------------------------------------------------------- .1234567891234567891234567891234567891235000000123456789.1234567891234567891234567891230000000000000000
123456789123.1234567891234567891234567891000000000000000000
#############################################################
SQL> col id for 9999999999999999999.99999999999999999999999999999999999999999999;
SP2-0246: Illegal FORMAT string "9999999999999999999.99999999999999999999999999999999999999999999"
SQL> col id for 9999999999999999999.9999999999999999999999999999999999999999
SQL> select * from test2;
ID
------------------------------------------------------------- .1234567891234567891234567891234567891235123456789.1234567891234567891234567891230000000000
123456789123.1234567891234567891234567891000000000000
12345678912345.1234567891234567891234567900000000000000
SQL>
这个问题纠结了很久，不明白为什么是39或40，后面在Oracle Database SQL Reference 10g Release 2终于找到解释了，如下所示：
p is the precision, or the total number of significant decimal digits, where the most
significant digit is the left-most nonzero digit, and the least significant digit is the
right-most known digit. Oracle guarantees the portability of numbers with
precision of up to 20 base-100 digits, which is equivalent to 39 or 40 decimal digits
depending on the position of the decimal point.
p是精度，或是有效十进制数的总位数。最大的有效数字是最左边的非零数字，而最小有效位是最右边的数字。 Oracle保证数字的可移植性
精度高达20 base-100 digits，相当于39位或40位十进制数字，取决于小数点的位置。