用Java解码一个字符串

时间:2011-03-23 16:30:31

标签: java urldecode

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

4 个答案:

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;
        }
    }
}