1>.NUMBER类型细讲：

Oracle number datatype 语法：NUMBER[(precision [, scale])]

简称：precision --> p

scale     --> s

NUMBER(p, s)

范围： 1 <= p <=38, -84 <= s <= 127

保存数据范围：-1.0e-130 <= number value < 1.0e+126

保存在机器内部的范围： 1 ~ 22 bytes

有效为：从左边第一个不为0的数算起的位数。

s的情况：

s > 0

精确到小数点右边s位，并四舍五入。然后检验有效位是否 <= p。

s < 0

精确到小数点左边s位，并四舍五入。然后检验有效位是否 <= p + |s|。

s = 0

此时NUMBER表示整数。

eg:

Actual Data   Specified As  Stored As

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

123.89           NUMBER         123.89

123.89           NUMBER(3)     124

123.89           NUMBER(6,2)   123.89

123.89           NUMBER(6,1)   123.9

123.89           NUMBER(4,2)   exceeds precision (有效位为5, 5 > 4)

123.89           NUMBER(6,-2)  100

.01234           NUMBER(4,5)   .01234 (有效位为4)

.00012           NUMBER(4,5)   .00012

.000127       NUMBER(4,5)   .00013

.0000012      NUMBER(2,7)   .0000012

.00000123     NUMBER(2,7)   .0000012

1.2e-4           NUMBER(2,5)   0.00012

1.2e-5           NUMBER(2,5)   0.00001

123.2564      NUMBER        123.2564

1234.9876     NUMBER(6,2)   1234.99

12345.12345   NUMBER(6,2)   Error (有效位为5+2 > 6)

1234.9876     NUMBER(6)     1235 (s没有表示s=0)

12345.345     NUMBER(5,-2)  12300

1234567       NUMBER(5,-2)  1234600

12345678      NUMBER(5,-2)  Error (有效位为8 > 7)

123456789     NUMBER(5,-4)  123460000

1234567890    NUMBER(5,-4)  Error (有效位为10 > 9)

12345.58      NUMBER(*, 1)  12345.6

0.1           NUMBER(4,5)   Error (0.10000, 有效位为5 > 4)

0.01234567    NUMBER(4,5)   0.01235

0.09999       NUMBER(4,5)   0.09999

2>.NUMBER在Oracle中如何存放？

先来做一个有趣的实验(可以将这两条语句放到文件中运行)：

建表numbers:

createtableNUMBERS

(

number1number(1),

negnumber1number(1),

number2number(2),

negnumber2number(2),

number3number(3),

negnumber3number(3),

number4number(4),

negnumber4number(4),

number5number(5),

negnumber5number(5),

number6number(6),

negnumber6number(6),

number7number(7),

negnumber7number(7),

number8number(8),

negnumber8number(8),

number9number(9),

negnumber9number(9),

number10number(10),

negnumber10number(10),

number11number(11),

negnumber11number(11),

number12number(12),

negnumber12number(12),

number13number(13),

negnumber13number(13),

number14number(14),

negnumber14number(14),

number15number(15),

negnumber15number(15),

number16number(16),

negnumber16number(16),

number17number(17),

negnumber17number(17),

number18number(18),

negnumber18number(18),

number19number(19),

negnumber19number(19),

number20number(20),

negnumber20number(20),

number21number(21),

negnumber21number(21),

number22number(22),

negnumber22number(22),

number23number(23),

negnumber23number(23),

number24number(24),

negnumber24number(24),

number25number(25),

negnumber25number(25),

number26number(26),

negnumber26number(26),

number27number(27),

negnumber27number(27),

number28number(28),

negnumber28number(28),

number29number(29),

negnumber29number(29),

number30number(30),

negnumber30number(30),

number31number(31),

negnumber31number(31),

umber32number(32),

negnumber32number(32),

number33number(33),

negnumber33number(33),

number34number(34),

negnumber34number(34),

number35number(35),

negnumber35number(35),

number36number(36),

negnumber36number(36),

number37number(37),

negnumber37number(37),

number38number(38),

negnumber38number(38)

);

插入一条数据：

insertintonumbersvalues(

9,-9,

99,-99,

999,-999,

9999,-9999,

99999,-99999,

999999,-999999,

9999999,-9999999,

99999999,-99999999,

999999999,-999999999,

9999999999,-9999999999,

99999999999,-99999999999,

999999999999,-999999999999,

9999999999999,-9999999999999,

99999999999999,-99999999999999,

999999999999999,-999999999999999,

9999999999999999,-9999999999999999,

99999999999999999,-99999999999999999,

999999999999999999,-999999999999999999,

9999999999999999999,-9999999999999999999,

99999999999999999999,-99999999999999999999,

999999999999999999999,-999999999999999999999,

9999999999999999999999,-9999999999999999999999,

99999999999999999999999,-99999999999999999999999,

999999999999999999999999,-999999999999999999999999,

9999999999999999999999999,-9999999999999999999999999,

99999999999999999999999999,-99999999999999999999999999,

999999999999999999999999999,-999999999999999999999999999,

9999999999999999999999999999,-9999999999999999999999999999,

99999999999999999999999999999,-99999999999999999999999999999,

999999999999999999999999999999,-999999999999999999999999999999,

9999999999999999999999999999999,-9999999999999999999999999999999,

99999999999999999999999999999999,-99999999999999999999999999999999,

999999999999999999999999999999999,-999999999999999999999999999999999,

9999999999999999999999999999999999,-9999999999999999999999999999999999,

99999999999999999999999999999999999,-99999999999999999999999999999999999,

999999999999999999999999999999999999,-999999999999999999999999999999999999,

9999999999999999999999999999999999999,-9999999999999999999999999999999999999,

99999999999999999999999999999999999999,-99999999999999999999999999999999999999);

现在来分析结果：

vsize函数为计算该字段在Oracle中存放占多少字节。

SQL>selectlength(number1),vsize(number1),length(negnumber1),vsize(negnumber1)fromnumbers;

LENGTH(NUMBER1)VSIZE(NUMBER1)LENGTH(NEGNUMBER1)VSIZE(NEGNUMBER1)--------------- -------------- ------------------ -----------------

1 2 2 3

SQL>selectlength(number38),vsize(number38),length(negnumber38),vsize(negnumber38)fromnumbers;

LENGTH(NUMBER38)VSIZE(NUMBER38)LENGTH(NEGNUMBER38)VSIZE(NEGNUMBER38)---------------- --------------- ------------------- ------------------

38 20 39 21

可以得出负数在Oracle中存放要比正数多占用一个字节的。

3>.NUMBER在Oracle中存储？

我们可以通过DUMP函数来转换数字的存储形式,一个简单的输出类似如下格式:

SQL> select dump(1) from dual;DUMP(1)

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

Typ=2 Len=2: 193,2

DUMP函数的输出格式类似:

类型 ，符号/指数位 [数字1，数字2，数字3，......，数字20]

各位的含义如下:

1.类型: Number型,Type=2 (类型代码可以从Oracle的文档上查到)

2.长度:指存储的字节数

3.符号/指数位

在存储上,Oracle对正数和负数分别进行存储转换:

正数：加1存储(为了避免Null)

负数：被101减,如果总长度小于21个字节，最后加一个102(是为了排序的需要)

指数位换算:

正数：指数=符号/指数位 - 193 (最高位为1是代表正数)

负数：指数=62 - 第一字节

4.从开始是有效的数据位

从开始是最高有效位,所存储的数值计算方法为：

将下面计算的结果加起来：

每个乘以100^(指数-N) (N是有效位数的顺序位，第一个有效位的N=0)

5. 举例说明

SQL> select dump(123456.789) from dual;DUMP(123456.789)

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

Typ=2 Len=6: 195,13,35,57,79,91

:   195 - 193 = 2

13 - 1    = 12 *100^(2-0) 120000

35 - 1    = 34 *100^(2-1) 3400

57 - 1    = 56 *100^(2-2) 56

79 - 1    = 78 *100^(2-3) .78

91 - 1    = 90 *100^(2-4) .009

123456.789

SQL> select dump(-123456.789) from dual;DUMP(-123456.789)

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

Typ=2 Len=7: 60,89,67,45,23,11,102

62 - 60 = 2(最高位是0，代表为负数)

101 - 89 = 12 *100^(2-0) 120000

101 - 67 = 34 *100^(2-1) 3400

101 - 45 = 56 *100^(2-2) 56

101 - 23 = 78 *100^(2-3) .78

101 - 11 = 90 *100^(2-4) .009

123456.789(-)

现在再考虑一下为什么在最后加102是为了排序的需要，-123456.789在数据库中实际存储为

60,89,67,45,23,11

而-123456.78901在数据库中实际存储为

60,89,67,45,23,11,91

可见，如果不在最后加上102，在排序时会出现-123456.789

对于2119号提问,第一个问题是:

1.请问为什么193,2各代表什么意思？

从上面就可以看到答案了.

2.还有NUMBER数字类型为什么有2个字节的长度呢？

对于这个问题,我想我们应该知道,所有数据类型最终在计算机里都以二进制存储,实际上所谓的数据类型都是我们定义的.所以存储只由算法决定.

所以这个问题是不成立的.比如:

SQL> select dump(110) from dual;DUMP(110)

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

Typ=2 Len=3: 194,2,11SQL> select dump(1100) from dual;DUMP(1100)

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

Typ=2 Len=2: 194,12

我们会看到,虽然1100>110,但是存储上1100却只占2字节,而110却占了3个字节.