oracle是堆屎山,Oracle NUMBER 类型细讲
1>.NUMBER類型細(xì)講：
Oracle number datatype 語法：NUMBER[(precision [, scale])]
簡稱：precision --> p
scale???? --> s
NUMBER(p, s)
范圍： 1 <= p <=38, -84 <= s <= 127
保存數(shù)據(jù)范圍：-1.0e-130 <= number value < 1.0e+126
保存在機(jī)器內(nèi)部的范圍： 1 ~ 22 bytes
有效為：從左邊第一個不為0的數(shù)算起的位數(shù)。
s的情況：
s > 0
精確到小數(shù)點(diǎn)右邊s位，并四舍五入。然后檢驗(yàn)有效位是否 <= p。
s < 0
精確到小數(shù)點(diǎn)左邊s位，并四舍五入。然后檢驗(yàn)有效位是否 <= p + |s|。
s = 0
此時NUMBER表示整數(shù)。
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 (有效位為5, 5 > 4)
123.89 ?? ?????? NUMBER(6,-2)? 100
.01234 ?? ?????? NUMBER(4,5)?? .01234 (有效位為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 (有效位為5+2 > 6)
1234.9876???? NUMBER(6)???? 1235 (s沒有表示s=0)
12345.345???? NUMBER(5,-2)? 12300
1234567?????? NUMBER(5,-2)? 1234600
12345678????? NUMBER(5,-2)? Error (有效位為8 > 7)
123456789???? NUMBER(5,-4)? 123460000
1234567890??? NUMBER(5,-4)? Error (有效位為10 > 9)
12345.58????? NUMBER(*, 1)? 12345.6
0.1?????????? NUMBER(4,5)?? Error (0.10000, 有效位為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？
先來做一個有趣的實(shí)驗(yàn)(可以將這兩條語句放到文件中運(yùn)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)
);
插入一條數(shù)據(jù)：
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);
現(xiàn)在來分析結(jié)果：
vsize函數(shù)為計算該字段在Oracle中存放占多少字節(jié)。
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
可以得出負(fù)數(shù)在Oracle中存放要比正數(shù)多占用一個字節(jié)的。
3>.NUMBER在Oracle中存儲？
我們可以通過DUMP函數(shù)來轉(zhuǎn)換數(shù)字的存儲形式,一個簡單的輸出類似如下格式:
SQL> select dump(1) from dual;DUMP(1)
------------------
Typ=2 Len=2: 193,2
DUMP函數(shù)的輸出格式類似:
類型 ，符號/指數(shù)位 [數(shù)字1，數(shù)字2，數(shù)字3，......，數(shù)字20]
各位的含義如下:
1.類型: Number型,Type=2 (類型代碼可以從Oracle的文檔上查到)
2.長度:指存儲的字節(jié)數(shù)
3.符號/指數(shù)位
在存儲上,Oracle對正數(shù)和負(fù)數(shù)分別進(jìn)行存儲轉(zhuǎn)換:
正數(shù)：加1存儲(為了避免Null)
負(fù)數(shù)：被101減,如果總長度小于21個字節(jié)，最后加一個102(是為了排序的需要)
指數(shù)位換算:
正數(shù)：指數(shù)=符號/指數(shù)位 - 193 (最高位為1是代表正數(shù))
負(fù)數(shù)：指數(shù)=62 - 第一字節(jié)
4.從開始是有效的數(shù)據(jù)位
從開始是最高有效位,所存儲的數(shù)值計算方法為：
將下面計算的結(jié)果加起來：
每個乘以100^(指數(shù)-N) (N是有效位數(shù)的順序位，第一個有效位的N=0)
5.?舉例說明
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，代表為負(fù)數(shù))
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(-)
現(xiàn)在再考慮一下為什么在最后加102是為了排序的需要，-123456.789在數(shù)據(jù)庫中實(shí)際存儲為
60,89,67,45,23,11
而-123456.78901在數(shù)據(jù)庫中實(shí)際存儲為
60,89,67,45,23,11,91
可見，如果不在最后加上102，在排序時會出現(xiàn)-123456.789
對于2119號提問,第一個問題是:
1.請問為什么193,2各代表什么意思？
從上面就可以看到答案了.
2.還有NUMBER數(shù)字類型為什么有2個字節(jié)的長度呢？
對于這個問題,我想我們應(yīng)該知道,所有數(shù)據(jù)類型最終在計算機(jī)里都以二進(jìn)制存儲,實(shí)際上所謂的數(shù)據(jù)類型都是我們定義的.所以存儲只由算法決定.
所以這個問題是不成立的.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
我們會看到,雖然1100>110,但是存儲上1100卻只占2字節(jié),而110卻占了3個字節(jié).
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