中文base64编码的问题！！！

jzywh 2005-05-14 11:21:58
我用这段函数加密中文，却解不出来？不知是什么原因？

<%
response.write base64Decode(base64Encode("我是中国人"))

%>


<%
'-------------------------------------------------
'函数功能：文字加密解密Base64算法
'加密函数：base64Encode(str) as String
'解密函数：base64Decode(str) as String
'-------------------------------------------------

const BASE_64_MAP_INIT = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/"
dim nl
' zero based arrays
dim Base64EncMap(63)
dim Base64DecMap(127)

' must be called before using anything else
PUBLIC SUB initCodecs()
' init vars
nl = "<P>" & chr(13) & chr(10)
' setup base 64
dim max, idx
max = len(BASE_64_MAP_INIT)
for idx = 0 to max - 1
' one based string
Base64EncMap(idx) = mid(BASE_64_MAP_INIT, idx + 1, 1)
next
for idx = 0 to max - 1
Base64DecMap(ASC(Base64EncMap(idx))) = idx
next
END SUB

' encode base 64 encoded string
PUBLIC FUNCTION base64Encode(plain)
if len(plain) = 0 then
base64Encode = ""
exit function
end if
dim ret, ndx, by3, first, second, third
by3 = (len(plain) \ 3) * 3
ndx = 1
do while ndx <= by3
first = asc(mid(plain, ndx+0, 1))
second = asc(mid(plain, ndx+1, 1))
third = asc(mid(plain, ndx+2, 1))
ret = ret & Base64EncMap((first \ 4) AND 63 )
ret = ret & Base64EncMap(((first * 16) AND 48) + ((second \ 16) AND 15 ) )
ret = ret & Base64EncMap(((second * 4) AND 60) + ((third \ 64) AND 3 ) )
ret = ret & Base64EncMap(third AND 63)
ndx = ndx + 3
loop
' check for stragglers
if by3 < len(plain) then
first = asc(mid(plain, ndx+0, 1))
ret = ret & Base64EncMap((first \ 4) AND 63 )
if (len(plain) MOD 3 ) = 2 then
second = asc(mid(plain, ndx+1, 1))
ret = ret & Base64EncMap(((first * 16) AND 48) + ((second \16) AND 15 ))
ret = ret & Base64EncMap(((second * 4) AND 60))
else
ret = ret & Base64EncMap((first * 16) AND 48)
ret = ret & "="
end if
ret = ret & "="
end if
base64Encode = ret
END FUNCTION

' decode base 64 encoded string
PUBLIC FUNCTION base64Decode(scrambled)
if len(scrambled) = 0 then
base64Decode = ""
exit function
end if
' ignore padding
dim realLen
realLen = len(scrambled)
do while mid(scrambled, realLen, 1) = "="
realLen = realLen - 1
loop
dim ret, ndx, by4, first, second, third, fourth
ret = ""
by4 = (realLen \ 4) * 4
ndx = 1
do while ndx <= by4
first = Base64DecMap(asc(mid(scrambled, ndx+0, 1)))
second = Base64DecMap(asc(mid(scrambled, ndx+1, 1)))
third = Base64DecMap(asc(mid(scrambled, ndx+2, 1)))
fourth = Base64DecMap(asc(mid(scrambled, ndx+3, 1)))
ret = ret & chr( ((first * 4) AND 255) + ((second \ 16) AND 3))
ret = ret & chr( ((second * 16) AND 255) + ((third \ 4) AND 15) )
ret = ret & chr( ((third * 64) AND 255) + (fourth AND 63) )
ndx = ndx + 4
loop
' check for stragglers, will be 2 or 3 characters
if ndx < realLen then
first = Base64DecMap(asc(mid(scrambled, ndx+0, 1)))
second = Base64DecMap(asc(mid(scrambled, ndx+1, 1)))
ret = ret & chr( ((first * 4) AND 255) + ((second \ 16) AND 3))
if realLen MOD 4 = 3 then
third = Base64DecMap(asc(mid(scrambled,ndx+2,1)))
ret = ret & chr( ((second * 16) AND 255) + ((third \ 4) AND 15) )
end if
end if
base64Decode = ret
END FUNCTION

' initialize
call initCodecs

' Testing code
' dim inp, encode
' inp = "1k2.3#4@5*6%7(8<9Q0x"
' encode = base64Encode(inp)
' response.write "Encoded value = " & encode & "<br/>"
' response.write "Decoded value = " & base64Decode(encode) & "<br/>"
%>

...全文
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supertj 2005-05-20
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没人回答了?
supertj 2005-05-18
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请问你的怎么用UrlEncode解决的啊?我也碰到类似的问题.教教我吧?

如果方便的,发份源码我吧?还有quoted-printable我也不会解,能教教我吗?

我的信箱,jacky.tang@psh.com.cn.非常非常感谢.
jzywh 2005-05-18
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谢谢，我用UrlEncode解决了
超级大笨狼 2005-05-16
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如孟子的代码。
孟子，web开发版的牛魔王。
孟子E章 2005-05-16
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<script Language="Javascript" runat=server>
var keyStr = "ABCDEFGHIJKLMNOP" +
"QRSTUVWXYZabcdef" +
"ghijklmnopqrstuv" +
"wxyz0123456789+/" +
"=";

function encode64(input) {
input = escape(input);
var output = "";
var chr1, chr2, chr3 = "";
var enc1, enc2, enc3, enc4 = "";
var i = 0;

do {
chr1 = input.charCodeAt(i++);
chr2 = input.charCodeAt(i++);
chr3 = input.charCodeAt(i++);

enc1 = chr1 >> 2;
enc2 = ((chr1 & 3) << 4) | (chr2 >> 4);
enc3 = ((chr2 & 15) << 2) | (chr3 >> 6);
enc4 = chr3 & 63;

if (isNaN(chr2)) {
enc3 = enc4 = 64;
} else if (isNaN(chr3)) {
enc4 = 64;
}

output = output +
keyStr.charAt(enc1) +
keyStr.charAt(enc2) +
keyStr.charAt(enc3) +
keyStr.charAt(enc4);
chr1 = chr2 = chr3 = "";
enc1 = enc2 = enc3 = enc4 = "";
} while (i < input.length);

return output;
}

function decode64(input) {
var output = "";
var chr1, chr2, chr3 = "";
var enc1, enc2, enc3, enc4 = "";
var i = 0;

// remove all characters that are not A-Z, a-z, 0-9, +, /, or =
var base64test = /[^A-Za-z0-9\+\/\=]/g;
if (base64test.exec(input)) {
alert("There were invalid base64 characters in the input text.\n" +
"Valid base64 characters are A-Z, a-z, 0-9, '+', '/', and '='\n" +
"Expect errors in decoding.");
}
input = input.replace(/[^A-Za-z0-9\+\/\=]/g, "");

do {
enc1 = keyStr.indexOf(input.charAt(i++));
enc2 = keyStr.indexOf(input.charAt(i++));
enc3 = keyStr.indexOf(input.charAt(i++));
enc4 = keyStr.indexOf(input.charAt(i++));

chr1 = (enc1 << 2) | (enc2 >> 4);
chr2 = ((enc2 & 15) << 4) | (enc3 >> 2);
chr3 = ((enc3 & 3) << 6) | enc4;

output = output + String.fromCharCode(chr1);

if (enc3 != 64) {
output = output + String.fromCharCode(chr2);
}
if (enc4 != 64) {
output = output + String.fromCharCode(chr3);
}

chr1 = chr2 = chr3 = "";
enc1 = enc2 = enc3 = enc4 = "";

} while (i < input.length);

return unescape(output);
}
Response.Write(("我是中国人"))
Response.Write(encode64("我是中国人"))
Response.Write(decode64(encode64("我是中国人")))
</script>


孟子E章 2005-05-16
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<script Language="Javascript" runat=server>
var keyStr = "ABCDEFGHIJKLMNOP" +
"QRSTUVWXYZabcdef" +
"ghijklmnopqrstuv" +
"wxyz0123456789+/" +
"=";

function encode64(input) {
input = escape(input);
var output = "";
var chr1, chr2, chr3 = "";
var enc1, enc2, enc3, enc4 = "";
var i = 0;

do {
chr1 = input.charCodeAt(i++);
chr2 = input.charCodeAt(i++);
chr3 = input.charCodeAt(i++);

enc1 = chr1 >> 2;
enc2 = ((chr1 & 3) << 4) | (chr2 >> 4);
enc3 = ((chr2 & 15) << 2) | (chr3 >> 6);
enc4 = chr3 & 63;

if (isNaN(chr2)) {
enc3 = enc4 = 64;
} else if (isNaN(chr3)) {
enc4 = 64;
}

output = output +
keyStr.charAt(enc1) +
keyStr.charAt(enc2) +
keyStr.charAt(enc3) +
keyStr.charAt(enc4);
chr1 = chr2 = chr3 = "";
enc1 = enc2 = enc3 = enc4 = "";
} while (i < input.length);

return output;
}

function decode64(input) {
var output = "";
var chr1, chr2, chr3 = "";
var enc1, enc2, enc3, enc4 = "";
var i = 0;

// remove all characters that are not A-Z, a-z, 0-9, +, /, or =
var base64test = /[^A-Za-z0-9\+\/\=]/g;
if (base64test.exec(input)) {
alert("There were invalid base64 characters in the input text.\n" +
"Valid base64 characters are A-Z, a-z, 0-9, '+', '/', and '='\n" +
"Expect errors in decoding.");
}
input = input.replace(/[^A-Za-z0-9\+\/\=]/g, "");

do {
enc1 = keyStr.indexOf(input.charAt(i++));
enc2 = keyStr.indexOf(input.charAt(i++));
enc3 = keyStr.indexOf(input.charAt(i++));
enc4 = keyStr.indexOf(input.charAt(i++));

chr1 = (enc1 << 2) | (enc2 >> 4);
chr2 = ((enc2 & 15) << 4) | (enc3 >> 2);
chr3 = ((enc3 & 3) << 6) | enc4;

output = output + String.fromCharCode(chr1);

if (enc3 != 64) {
output = output + String.fromCharCode(chr2);
}
if (enc4 != 64) {
output = output + String.fromCharCode(chr3);
}

chr1 = chr2 = chr3 = "";
enc1 = enc2 = enc3 = enc4 = "";

} while (i < input.length);

return unescape(output);
}
Response.Write(("我是中国人"))
Response.Write(encode64("我是中国人"))
Response.Write(decode64(encode64("我是中国人")))
</script>


笨笨兔兔兔兔兔 2005-05-16
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加密

GB2312编码 -> UTF8


解密


UTF8 -> GB2312编码

就是说需要2个转换函数


看看笨狼的blog，里面也许有你要的东西
jzywh 2005-05-16
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不行啊，用 UTF-8编码之后再base64编码，得出的字符串不满足 base64的格式，而且解码出来是空的

<!--#include virtual="/inc/security.asp"-->
<!--#include virtual="/inc/base64.asp"-->
<%
dim tempStr
tempStr = "werwerwrw你好啊!你好啊!你好啊!你好啊!"
tempStr = utf8Encode(tempStr)
response.write tempStr & "<br>"
tempStr = base64Encode(base64Encode)
response.write tempStr & "<br>"
%>
是是非非 2005-05-14
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是的

最好是你修改一下上面的Base64编码/解码函数
这样效率高的多
可以省去两次字符歘扫描
jzywh 2005-05-14
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加密

GB2312编码 -> UTF8


解密


UTF8 -> GB2312编码

就是说需要2个转换函数
raas 2005-05-14
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学习
是是非非 2005-05-14
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上面用的都是 Mid Chr Asc等不带B的函数

…………呵呵，当然不行了
这有一个UTF-8编码倒GB2312编码的转换函数

Public Function Bytes2BSTR(v)
Dim r,i,t,n : r = ""
For i = 1 To LenB(v)
t = AscB(MidB(v,i,1))
If t < &H80 Then
r = r & Chr(t)
Else
n = AscB(MidB(v,i+1,1))
r = r & Chr(CLng(t) * &H100 + CInt(n))
i = i + 1
End If
Next
Bytes2BSTR = r
End Function

＝＝＝＝＝＝＝＝＝
从GB2312到UTF-8的也不难了
是是非非 2005-05-14
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上面的64Base是基于字节编码的

中文是双字节的

所以要先转换一下，把中文字符串转换成UF-8编码的字符串
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NULL 博文链接：https://onestopweb.iteye.com/blog/2379876
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