如何正确解码Java中的以下字符串
http%3A//www.google.ru/search%3Fhl%3Dru%26q%3Dla+mer+powder%26btnG%3D%u0420%A0%u0421%u045F%u0420%A0%u0421%u2022%u0420%A0%u0421%u2018%u0420%u040E%u0420%u0453%u0420%A0%u0421%u201D+%u0420%A0%u0420%u2020+Google%26lr%3D%26rlz%3D1I7SKPT_ru
当我使用URLDecoder.decode()时出现以下错误
java.lang.IllegalArgumentException: URLDecoder: Illegal hex characters in escape (%) pattern - For input string: "u0"
谢谢, 戴夫
答案 0 :(得分:2)
根据Wikipedia,“Unicode字符存在非标准编码:%uxxxx
,其中xxxx
是Unicode值”。
继续:“此行为未由任何RFC指定,并已被W3C拒绝”。
您的URL包含此类令牌,而Java URLDecoder实现不支持这些令牌。
答案 1 :(得分:2)
%uXXXX
编码是非标准的,实际上被W3C拒绝了,所以很自然,URLDecoder不理解它。
您可以制作小功能,通过在编码字符串中将%uXXYY
替换为%XX%YY
来修复它。然后你可以正常处理和解码固定的字符串。
答案 2 :(得分:1)
我们从Vartec的解决方案开始,但发现了其他问题。此解决方案适用于UTF-16,但可以更改为返回UTF-8。为清晰起见,请保留替换全部内容,您可以在http://www.cogniteam.com/wiki/index.php?title=DecodeEncodeJavaScript
阅读更多内容static public String unescape(String escaped) throws UnsupportedEncodingException
{
// This code is needed so that the UTF-16 won't be malformed
String str = escaped.replaceAll("%0", "%u000");
str = str.replaceAll("%1", "%u001");
str = str.replaceAll("%2", "%u002");
str = str.replaceAll("%3", "%u003");
str = str.replaceAll("%4", "%u004");
str = str.replaceAll("%5", "%u005");
str = str.replaceAll("%6", "%u006");
str = str.replaceAll("%7", "%u007");
str = str.replaceAll("%8", "%u008");
str = str.replaceAll("%9", "%u009");
str = str.replaceAll("%A", "%u00A");
str = str.replaceAll("%B", "%u00B");
str = str.replaceAll("%C", "%u00C");
str = str.replaceAll("%D", "%u00D");
str = str.replaceAll("%E", "%u00E");
str = str.replaceAll("%F", "%u00F");
// Here we split the 4 byte to 2 byte, so that decode won't fail
String [] arr = str.split("%u");
Vector<String> vec = new Vector<String>();
if(!arr[0].isEmpty())
{
vec.add(arr[0]);
}
for (int i = 1 ; i < arr.length ; i++) {
if(!arr[i].isEmpty())
{
vec.add("%"+arr[i].substring(0, 2));
vec.add("%"+arr[i].substring(2));
}
}
str = "";
for (String string : vec) {
str += string;
}
// Here we return the decoded string
return URLDecoder.decode(str,"UTF-16");
}
答案 3 :(得分:1)
在仔细研究了@ariy提供的解决方案之后,我创建了一个基于Java的解决方案,该解决方案还可以对已经被分成两部分的编码字符(即缺少编码字符的一半)具有弹性。这发生在我的用例中,我需要解码有时在2000个字符长度切断的长网址。见What is the maximum length of a URL in different browsers?
public class Utils {
private static Pattern validStandard = Pattern.compile("%([0-9A-Fa-f]{2})");
private static Pattern choppedStandard = Pattern.compile("%[0-9A-Fa-f]{0,1}$");
private static Pattern validNonStandard = Pattern.compile("%u([0-9A-Fa-f][0-9A-Fa-f])([0-9A-Fa-f][0-9A-Fa-f])");
private static Pattern choppedNonStandard = Pattern.compile("%u[0-9A-Fa-f]{0,3}$");
public static String resilientUrlDecode(String input) {
String cookedInput = input;
if (cookedInput.indexOf('%') > -1) {
// Transform all existing UTF-8 standard into UTF-16 standard.
cookedInput = validStandard.matcher(cookedInput).replaceAll("%00%$1");
// Discard chopped encoded char at the end of the line (there is no way to know what it was)
cookedInput = choppedStandard.matcher(cookedInput).replaceAll("");
// Handle non standard (rejected by W3C) encoding that is used anyway by some
// See: https://stackoverflow.com/a/5408655/114196
if (cookedInput.contains("%u")) {
// Transform all existing non standard into UTF-16 standard.
cookedInput = validNonStandard.matcher(cookedInput).replaceAll("%$1%$2");
// Discard chopped encoded char at the end of the line
cookedInput = choppedNonStandard.matcher(cookedInput).replaceAll("");
}
}
try {
return URLDecoder.decode(cookedInput,"UTF-16");
} catch (UnsupportedEncodingException e) {
// Will never happen because the encoding is hardcoded
return null;
}
}
}