Question

illegal hex characters in escape pattern什么错误

Answer 1

如果收到的HTTP请求参歼扒数中有一个字符串，是中文，比如“10%是黄段子”，服务器段使用URLDecoder.decode就会出现此异常。
URL只能使用英文字母、阿拉伯数字和某些标点符号，不能使用其他文字和符号。如果内容中存在中文，必须要进行编解码。“10%是黄段子”转码过后是“10%25%E6%98%AF%E9%BB%84%E6%AE%B5%E5%AD%90”%被用来作为转义字符使用。可以看URLDecoder.decode源码：
public static String decode(String s, String enc)
throws UnsupportedEncodingException{

boolean needToChange = false;
int numChars = s.length();
StringBuffer sb = new StringBuffer(numChars > 500 ? numChars / 2 : numChars);
int i = 0;

if (enc.length() == 0) {
throw new UnsupportedEncodingException ("URLDecoder: empty string enc parameter");
}

char c;
byte[] bytes = null;
while (i < numChars) {
c = s.charAt(i);
switch (c) {
case '+':
sb.append(' ');
i++;
needToChange = true;
break;
case '%':
/*
* Starting with this instance of %, process all
* consecutive substrings of the form %xy. Each
* substring %xy will yield a byte. Convert all
* consecutive bytes obtained this way to whatever
* character(s) they represent in the provided
* encoding.
*/

try {

// (numChars-i)/3 is an upper bound for the number
// of remaining bytes
if (bytes == null)
bytes = new byte[(numChars-i)/3];
int pos = 0;

while ( ((i+2) <雹伍 numChars) &&
(c=='%')) {
int v = Integer.parseInt(s.substring(i+1,i+3),16);
if (v < 0)
throw new IllegalArgumentException("URLDecoder: Illegal hex characters in escape (%) pattern - negative value");
bytes[pos++] = (byte) v;
i+= 3;
if (i < numChars)
c = s.charAt(i);
}

// A trailing, incomplete byte encoding such as
// "%x" will cause an exception to be thrown

if ((i < numChars) && (c=='%'))
throw new IllegalArgumentException(
"氏肆昌URLDecoder: Incomplete trailing escape (%) pattern");

sb.append(new String(bytes, 0, pos, enc));
} catch (NumberFormatException e) {
throw new IllegalArgumentException(
"URLDecoder: Illegal hex characters in escape (%) pattern - "
+ e.getMessage());
}
needToChange = true;
break;
default:
sb.append(c);
i++;
break;
}
}

return (needToChange? sb.toString() : s);
}

上面的字符串中'%'是一个中文字符'是'，而转换的实现是将%后面的两个字符一起转为一个16进制数。拿"%是"来转换数字，肯定会有NumberFormatException异常。
类似的如果请求字符串中有'+'，也会有问题。因为'+'被当做空格使用了。
一个解决办法就是将%替换为%25。
public static String replacer(StringBuffer outBuffer) {
String data = outBuffer.toString();
try {
data = data.replaceAll("%(?![0-9a-fA-F]{2})", "%25");
data = data.replaceAll("\\+", "%2B");
data = URLDecoder.decode(data, "utf-8");
} catch (Exception e) {
e.printStackTrace();
}
return data;
}

这里使用了一个特殊正则表达式：零宽负向先行断言(zero-width negative lookahead assertion)，模式为(?!pattern)，代表字符串中的一个位置，紧接该位置之后的字符序列不能匹配pattern。%(?![0-9a-fA-F]{2})意思是'%'开始，但是后面两个字符不是数字，也不是字母。