[20200103]GUID轉換GUID_BASE64.txt
--//最近在做一個專案優化,里面大量使用uuid.優缺點在鏈接:http://blog.itpub.net.x.y265/viewspace-2670513/=>[20191225]主鍵使
--//用uuid優缺點.txt 有相關討論.我自己的觀點不要濫用,或者講到處都用,合理使用才是比較正確的選擇.
--//昨天看12c相關書籍,發現oracle給每個PDB設定唯一GUID.我查看視圖V$CONTAINERs,發現有1個欄位GUID_BASE64,很明顯這個從guid轉
--//換過來,自己想知道這個轉換如何實作的?
1.環境:
[email protected]:1521/orclcdb> select banner from v$version;
BANNER
----------------------------------------------------------------------
Oracle Database 18c Enterprise Edition Release 18.0.0.0.0 - Production
[email protected]:1521/orclcdb> select sys_guid() from dual ;
SYS_GUID()
--------------------------------
9B25CF226E3E36A5E0558253DD747177
[email protected]:1521/orclcdb> select CON_ID,DBID,CON_UID,NAME,GUID,GUID_BASE64 from V$CONTAINERs;
CON_ID DBID CON_UID NAME GUID GUID_BASE64
------ ---------- ---------- -------- -------------------------------- ------------------------
1 2756091850 1 CDB$ROOT 64A52F53A7683286E053CDA9E80AED76 ZKUvU6doMobgU82p6Artdg==
2 1474312904 1474312904 PDB$SEED 742DCFA2CE044FDEE0558253DD747177 dC3Pos4ET97gVYJT3XRxdw==
3 115310104 115310104 ORCL 74A69DC145F5662BE0558253DD747177 dKadwUX1ZivgVYJT3XRxdw==
--//注意看sys_guid()后面16位E0558253DD747177,居然沒有變化,不知道為什么.
--//GUID_BASE64后面2位是字符'==',不可能3個正好都是==,一定是用來填充保持字串長度24.
2.首先我必須確定GUID_BASE64的編碼:
--//base64,我的理解就是64進制,確定編碼很重要,最容易聯想到的rowid編碼也是64進制,是否其編碼與它一樣.
--//我檢索發現如下鏈接:
https://docs.oracle.com/cd/E18150_01/javadocs/DevelopmentKit/com/stc/connector/framework/util/Base64.html
--//內容如下:
Table 1: The Base64 Alphabet
Value Encoding Value Encoding Value Encoding Value Encoding
0 A 17 R 34 i 51 z
1 B 18 S 35 j 52 0
2 C 19 T 36 k 53 1
3 D 20 U 37 l 54 2
4 E 21 V 38 m 55 3
5 F 22 W 39 n 56 4
6 G 23 X 40 o 57 5
7 H 24 Y 41 p 58 6
8 I 25 Z 42 q 59 7
9 J 26 a 43 r 60 8
10 K 27 b 44 s 61 9
11 L 28 c 45 t 62 +
12 M 29 d 46 u 63 /
13 N 30 e 47 v
14 O 31 f 48 w (pad) =
15 P 32 g 49 x
16 Q 33 h 50 y
--//= 作為pad與看到結果一樣.
The encoded output stream must be represented in lines of no more than 76 characters each. All line breaks or other
characters no found in Table 1 must be ignored by decoding software. In base64 data, characters other than those in
Table 1, line breaks, and other white space probably indicate a transmission error, about which a warning message or
even a message rejection might be appropriate under some circumstances.
編碼的輸出流必須以不超過76個字符的行表示，所有行打斷或其他解碼軟體必須忽略表1中沒有找到的字符，在base64資料中，除了那些
表1、斷行和其他空白可能表示傳輸錯誤，有關該錯誤的警告訊息或在某些情況下，即使是拒絕資訊也可能是合適的，
Special processing is performed if fewer than 24 bits are available at the end of the data being encoded. A full
encoding quantum is always completed at the end of a body. When fewer than 24 input bits are available in an input
group, zero bits are added (on the right) to form an integral number of 6-bit groups. Padding at the end of the data is
performed using the "=" character. Since all base64 input is an integral number of octets, only the following cases can
arise: (1) the final quantum of encoding input is an integral multiple of 24 bits; here, the final unit of encoded
output will be an integral multiple of 4 characters with no "=" padding, (2) the final quantum of encoding input is
exactly 8 bits; here, the final unit of encoded output will be two characters followed by two "=" padding characters, or
(3) the final quantum of encoding input is exactly 16 bits; here, the final unit of encoded output will be three
characters followed by one "=" padding character.
如果在被編碼資料的末尾有少于24位可用，則執行特殊處理，一個完整的編碼量子總是在物體的末端完成，當一個輸入中可用的輸入位少
于24位時組，零位被添加(在右邊)形成一個6位組的整數，資料末尾的填充物是使用"="字符執行，由于所有base64輸入都是八進制的整數
，所以只有以下情況才能產生：
(1)編碼輸入的最終量子是24位的整數倍；這里是編碼的最終單位輸出將是4個字符的整數倍，沒有"="填充.
(2)編碼輸入的最終數量是8位，編碼輸出的最終單位將是兩個字符，后面跟著兩個"="填充字符，或者
(3)編碼輸入的最終量子正好是16位；這里，編碼輸出的最終單位是三位字符后面跟著一個"="填充字符，
Because it is used only for padding at the end of the data, the occurrence of any "=" characters may be taken as
evidence that the end of the data has been reached (without truncation in transit). No such assurance is possible,
however, when the number of octets transmitted was a multiple of three and no "=" characters are present.
因為它只用于資料末尾的填充，所以任何"="字符的出現都可以作為資料結束的證據已經到達(在運輸程序中沒有截斷)，不可能有這樣的
保證，然而，當傳輸的八位數是三個的倍數，并且沒有"="字符存在時，
Any characters outside of the base64 alphabet are to be ignored in base64-encoded data.
在base64編碼的資料中，base64字母表之外的任何字符都將被忽略，
Care must be taken to use the proper octets for line breaks if base64 encoding is applied directly to text material that
has not been converted to canonical form. In particular, text line breaks must be converted into CRLF sequences prior to
base64 encoding. The important thing to note is that this may be done directly by the encoder rather than in a prior
canonicalization step in some implementations.
如果base64編碼直接應用于尚未轉換為規范形式，特別是，文本換行必須先轉換成CRLF序列，base64編碼，重要的是要注意的是，這可能
是由編碼器直接完成的，而不是在以前完成的，規范化步驟在一些實作中，
NOTE: There is no need to worry about quoting potential boundary delimiters within base64-encoded bodies within
multipart entities because no hyphen characters are used in the base64 encoding.
--//注:翻譯我使用金山詞霸,可能存在一些瑕疵...
--//這樣base64的編碼可以確定.A_Z a-z 0-9 +/
3.我寫一個測驗腳本:
$ cat 64base.sh
#! /bin/bash
v2=$1
BASE64=($( echo {A..Z} {a..z} {0..9} + / ))
res=''
for i in $(echo "obase=64;ibase=16; $v2" | bc| tr -d '\\\r\n')
do
res=${res}${BASE64[$(( 10#$i ))]}
done
echo $res
$ ./64base.sh 74A69DC145F5662BE0558253DD747177
B0pp3BRfVmK+BVglPddHF3
--//完全對不上.我再仔細看鏈接https://docs.oracle.com/cd/E18150_01/javadocs/DevelopmentKit/com/stc/connector/framework/util/Base64.html
--//說明.
When fewer than 24 input bits are available in an input group, zero bits are added (on the right) to form an integral
number of 6-bit groups. Padding at the end of the data is performed using the "=" character.
--//有1個非常明顯的提示 "zero bits are added (on the right) to form an integral number of 6-bit groups".
--//guid的顯示74A69DC145F5662BE0558253DD747177,占32字符.32*4 = 128 bits.
--//base64 相當于2^6,也就是6 bits表示1個64進制字符.
--//128/6 = 21.333,明顯無法整除,這樣結尾要補上1個0(占4bits).
--//(128+4)/6 = 22,這樣正好整除.
--//補上1個0再計算如下:
$ ./64base.sh 74A69DC145F5662BE0558253DD7471770
dKadwUX1ZivgVYJT3XRxdw
--//^_^正好對上.可以驗證看看.
$ echo 64A52F53A7683286E053CDA9E80AED760 742DCFA2CE044FDEE0558253DD7471770 74A69DC145F5662BE0558253DD7471770 | tr ' ' '\n' | xargs -IQ ./64base.sh Q
ZKUvU6doMobgU82p6Artdg
dC3Pos4ET97gVYJT3XRxdw
dKadwUX1ZivgVYJT3XRxdw
[email protected]:1521/orclcdb> select CON_ID,DBID,CON_UID,NAME,GUID,GUID_BASE64 from V$CONTAINERs;
CON_ID DBID CON_UID NAME GUID GUID_BASE64
------ ---------- ---------- -------- -------------------------------- ------------------------
1 2756091850 1 CDB$ROOT 64A52F53A7683286E053CDA9E80AED76 ZKUvU6doMobgU82p6Artdg==
2 1474312904 1474312904 PDB$SEED 742DCFA2CE044FDEE0558253DD747177 dC3Pos4ET97gVYJT3XRxdw==
3 115310104 115310104 ORCL 74A69DC145F5662BE0558253DD747177 dKadwUX1ZivgVYJT3XRxdw==
--//對比完全能對上.當然不包括后面的兩個=.
4.修改腳本如下:
$ cat o64base.sh
#! /bin/bash
# convert guid to guid_base64
odebug=${ODEBUG:-0}
v2=${1}0
BASE64=($( echo {A..Z} {a..z} {0..9} + / ))
res=''
for i in $(echo "obase=64;ibase=16; $v2" | bc| tr -d '\\\r\n')
do
res=${res}${BASE64[$(( 10#$i ))]}
done
if [ $odebug -eq 1 ] ; then
echo v2=$v2 res=$res
fi
res=${res}==
echo $res
$ ./o64base.sh 74A69DC145F5662BE0558253DD747177
dKadwUX1ZivgVYJT3XRxdw==
5.總結:
--//純屬無聊,浪費一個下午探究這個問題.
--//在測驗時我使用鏈接https://toolslick.com/conversion/data/guid的在線轉換工具,不然估計我無法猜測到如何實作.
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標籤：Oracle