javascript高手，救命啊!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
碰到ü之类的特殊字符怎么解？我在网上看到别人用的代码可以解开一般的字母，但遇到特殊字符就不行了。怎么办？！！～～～
例如：
ü加密后是：
#@~^AQAAAA==üAAAAAA==^#~@
但就不能解密还原成ü
我在网上看到别人的解密函数，但不能解
function toEncode(s,l)
{enc=new ActiveXObject("Scripting.Encoder");
return enc.EncodeScriptFile("."+l,s,0,l+"cript");
}
var STATE_COPY_INPUT = 100
var STATE_READLEN = 101
var STATE_DECODE = 102
var STATE_UNESCAPE = 103
var pick_encoding = new Array(
1, 2, 0, 1, 2, 0, 2, 0, 0, 2, 0, 2, 1, 0, 2, 0,
1, 0, 2, 0, 1, 1, 2, 0, 0, 2, 1, 0, 2, 0, 0, 2,
1, 1, 0, 2, 0, 2, 0, 1, 0, 1, 1, 2, 0, 1, 0, 2,
1, 0, 2, 0, 1, 1, 2, 0, 0, 1, 1, 2, 0, 1, 0, 2
)
var rawData = new Array(
0x64,0x37,0x69, 0x50,0x7E,0x2C, 0x22,0x5A,0x65, 0x4A,0x45,0x72,
0x61,0x3A,0x5B, 0x5E,0x79,0x66, 0x5D,0x59,0x75, 0x5B,0x27,0x4C,
0x42,0x76,0x45, 0x60,0x63,0x76, 0x23,0x62,0x2A, 0x65,0x4D,0x43,
0x5F,0x51,0x33, 0x7E,0x53,0x42, 0x4F,0x52,0x20, 0x52,0x20,0x63,
0x7A,0x26,0x4A, 0x21,0x54,0x5A, 0x46,0x71,0x38, 0x20,0x2B,0x79,
0x26,0x66,0x32, 0x63,0x2A,0x57, 0x2A,0x58,0x6C, 0x76,0x7F,0x2B,
0x47,0x7B,0x46, 0x25,0x30,0x52, 0x2C,0x31,0x4F, 0x29,0x6C,0x3D,
0x69,0x49,0x70, 0x3F,0x3F,0x3F, 0x27,0x78,0x7B, 0x3F,0x3F,0x3F,
0x67,0x5F,0x51, 0x3F,0x3F,0x3F, 0x62,0x29,0x7A, 0x41,0x24,0x7E,
0x5A,0x2F,0x3B, 0x66,0x39,0x47, 0x32,0x33,0x41, 0x73,0x6F,0x77,
0x4D,0x21,0x56, 0x43,0x75,0x5F, 0x71,0x28,0x26, 0x39,0x42,0x78,
0x7C,0x46,0x6E, 0x53,0x4A,0x64, 0x48,0x5C,0x74, 0x31,0x48,0x67,
0x72,0x36,0x7D, 0x6E,0x4B,0x68, 0x70,0x7D,0x35, 0x49,0x5D,0x22,
0x3F,0x6A,0x55, 0x4B,0x50,0x3A, 0x6A,0x69,0x60, 0x2E,0x23,0x6A,
0x7F,0x09,0x71, 0x28,0x70,0x6F, 0x35,0x65,0x49, 0x7D,0x74,0x5C,
0x24,0x2C,0x5D, 0x2D,0x77,0x27, 0x54,0x44,0x59, 0x37,0x3F,0x25,
0x7B,0x6D,0x7C, 0x3D,0x7C,0x23, 0x6C,0x43,0x6D, 0x34,0x38,0x28,
0x6D,0x5E,0x31, 0x4E,0x5B,0x39, 0x2B,0x6E,0x7F, 0x30,0x57,0x36,
0x6F,0x4C,0x54, 0x74,0x34,0x34, 0x6B,0x72,0x62, 0x4C,0x25,0x4E,
0x33,0x56,0x30, 0x56,0x73,0x5E, 0x3A,0x68,0x73, 0x78,0x55,0x09,
0x57,0x47,0x4B, 0x77,0x32,0x61, 0x3B,0x35,0x24, 0x44,0x2E,0x4D,
0x2F,0x64,0x6B, 0x59,0x4F,0x44, 0x45,0x3B,0x21, 0x5C,0x2D,0x37,
0x68,0x41,0x53, 0x36,0x61,0x58, 0x58,0x7A,0x48, 0x79,0x22,0x2E,
0x09,0x60,0x50, 0x75,0x6B,0x2D, 0x38,0x4E,0x29, 0x55,0x3D,0x3F
)
var transformed = new Array()
for (var i=0; i<3; i++) transformed[i] = new Array()
for (var i=31; i<=126; i++) for (var j=0; j<3; j++) transformed[j][rawData[(i-31) * 3 + j]] = (i==31) ? 9 : i
var digits = new Array()
for (var i=0; i<26; i++)
{
digits["A".charCodeAt(0)+i] = i
digits["a".charCodeAt(0)+i] = i+26
}
for (var i=0; i<10; i++) digits["0".charCodeAt(0)+i] = i+52
digits[0x2b] = 62
digits[0x2f] = 63
function unescape(char)
{
var escapes = "#&!*$"
var escaped = "\r\n<>@"
if (char.charCodeAt(0) > 126) return char
if (escapes.indexOf(char) != -1) return escaped.substr(escapes.indexOf(char), 1)
return "?"
}
function decodeBase64(string)
{
var val = 0
val += (digits[string.substr(0,1).charCodeAt(0)] << 2)
val += (digits[string.substr(1,1).charCodeAt(0)] >> 4)
val += (digits[string.substr(1,1).charCodeAt(0)] & 0xf) << 12
val += ((digits[string.substr(2,1).charCodeAt(0)] >> 2) << 8)
val += ((digits[string.substr(2,1).charCodeAt(0)] & 0x3) << 22)
val += (digits[string.substr(3,1).charCodeAt(0)] << 16)
return val
}
function tranEncode(encodingString)
{
var marker = "#@~^"
var stringIndex = 0
var scriptIndex = -1
var unEncodingIndex = 0
var char = null
var encodingLength = unEncodinglength = 0
var state = STATE_COPY_INPUT
var unEncodingString = ""
var re, arr
while(state)
{
switch (state)
{
case (STATE_COPY_INPUT) :
scriptIndex = encodingString.indexOf(marker, stringIndex)
if (scriptIndex != -1)
{
unEncodingString += encodingString.substring(stringIndex, scriptIndex)
scriptIndex += marker.length
state = STATE_READLEN
}
else
{
stringIndex = stringIndex==0 ? 0 : stringIndex
unEncodingString += encodingString.substr(stringIndex, encodingString.length)
state = 0
}
break
case (STATE_READLEN) :
encodingLength = encodingString.substr(scriptIndex, 6)
unEncodinglength = decodeBase64(encodingLength)
scriptIndex += (6 + "==".length)
state = STATE_DECODE
break
case (STATE_DECODE) :
if (!unEncodinglength)
{
stringIndex = scriptIndex + "DQgAAA==^#~@".length
unEncodingIndex = 0
state = STATE_COPY_INPUT
break
}
char = encodingString.substr(scriptIndex, 1)
if (char == "@") state = STATE_UNESCAPE
else
{
if (char.charCodeAt(0) < 0xFF)
{
unEncodingString += String.fromCharCode(transformed[pick_encoding[unEncodingIndex%64]][char.charCodeAt(0)])
unEncodingIndex++
}
else
{
unEncodingString += char
}
scriptIndex++
unEncodinglength--
break
}
case STATE_UNESCAPE:
unEncodingString += unescape(encodingString.substr(++scriptIndex, 1))
scriptIndex++; unEncodinglength -=2
unEncodingIndex++
state = STATE_DECODE
break
}
}
re = new RegExp("(JScript|VBscript).encode", "gmi")
while(arr = re.exec(unEncodingString)) unEncodingString = RegExp.leftContext + RegExp.$1 + RegExp.rightContext
return unEncodingString
}
问题点数：0、回复次数：2Top
1 楼cuixiping（无心●愚公）回复于 2004-12-04 09:07:30 得分 0
怎么个不行？可以的啊。Top
2 楼flyskytoday（夜漫漫路漫漫）回复于 2004-12-04 10:00:52 得分 0
加密后地函数你必须完整地列出
要是变动一点点都会无法解密Top
