2

如何在 Java 中正确解码以下字符串

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() 我出现以下错误

java.lang.IllegalArgumentException: URLDecoder: Illegal hex characters in escape (%) pattern - For input string: "u0"

谢谢,戴夫

4

4 回答 4

2

根据Wikipedia的说法,“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,已被 W3C 拒绝”。

您的 URL 包含此类标记,而 Java URLDecoder 实现不支持这些标记。

于 2011-03-23T16:36:45.220 回答
2

%uXXXXencoding 是非标准的,实际上被 W3C 拒绝了,所以 URLDecoder 不理解它是很自然的。

您可以制作小函数,通过在编码字符串中替换每次出现的%uXXYYwith来修复它。%XX%YY然后您可以正常处理和解码固定字符串。

于 2011-03-23T16:39:00.073 回答
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。为清楚起见,全部替换,您可以在http://www.cognitteam.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");
}
于 2011-07-25T09:22:59.223 回答
1

在仔细查看了@ariy 提出的解决方案之后,我创建了一个基于Java 的解决方案,该解决方案还可以抵抗被分成两部分的编码字符(即缺少一半的编码字符)。这发生在我的用例中,我需要解码有时会以 2000 个字符长度截断的长 url。请参阅不同浏览器中 URL 的最大长度是多少?

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;
        }
    }
}
于 2014-05-02T12:40:20.630 回答