2

hexstringをbase64に変換するのを手伝ってください

これが私が例外を取得しているcedeです

 String hexString = "bf940165bcc3bca12321a5cc4c753220129337b48ad129d880f718d147a2cd1bfa79de92239ef1bc06c2f05886b0cd5d";

private static String ConvertHexStringToBase64(String hexString) {
    System.out.println(hexString);
    if ((hexString.length()) % 2 > 0)
        throw new NumberFormatException("Input string was not in a correct format.");
     byte[] buffer = new byte[hexString.length() / 2];
        int i = 0;
        while (i < hexString.length())
        {
            buffer[i / 2] = Byte.parseByte(hexString.substring(i, 2));
            i += 2;
        }
        System.out.println("hexSring"+hexString+"afterconverttobase64"+Base64.encodeBase64String(buffer));
        return Base64.encodeBase64String(buffer);

}

ここで例外が発生します::bf940165bcc3bca12321a5cc4c753220129337b48ad129d880f718d147a2cd1bfa79de92239ef1bc06c2f05886b0cd5d Exception in thread "main" java.lang.NumberFormatException: For input string: "bf" at java.lang.NumberFormatException.forInputString(NumberFormatException.java:48) at java.lang.Integer.parseInt(Integer.java:449) at java.lang.Byte.parseByte(Byte.java:151) at java.lang.Byte.parseByte(Byte.java:108) at com.motorola.gst.DecryptTest3.ConvertHexStringToBase64(DecryptTest3.java:38) at com.motorola.gst.DecryptTest3.main(DecryptTest3.java:16)

4

2 に答える 2

4

First of all you have to specify the specify the radix(16 in your case) in the parseByte method to avoid the numberFormat exception :

 buffer[i / 2] = Byte.parseByte(hexString.substring(i, 2),16);

However your code seems broken, take a look at the corrected one : 

     if ((hexString.length()) % 2 > 0)
          throw new NumberFormatException("Input string was not in a correct format.");
       int[] buffer = new int[hexString.length() / 2];
          int i = 2;
          while (i < hexString.length())
          {
              buffer[i / 2] = Integer.parseInt(hexString.substring(i, i + 2),16);
              i += 2;
          }

Your loop was wrong and you have to parse as Integer because you have some values inside your input string that overflow the byte capability ...

If you need byte you could cast the parsed int to byte in this way : 

       byte[] buffer = new byte[hexString.length() / 2];
          int i = 2;
          while (i < hexString.length())
          {
              buffer[i / 2] = (byte)Integer.parseInt(hexString.substring(i, i + 2),16);
              i += 2;
          }
于 2012-10-13T06:57:21.640 に答える
-1

同様の解決策が見つかりました。共有するのが良いかもしれないと考えました：

    public static string convertHexToBase64String(String hexString)
    {
        string base64 = "";

        //--Important: remove "0x" groups from hexidecimal string--
        hexString = hexString.Replace("0x", "");

        byte[] buffer = new byte[hexString.Length / 2];

        for (int i = 0; i < hexString.Length; i++)
        {
            try
            {
                buffer[i / 2] = Convert.ToByte(Convert.ToInt32(hexString.Substring(i, 2), 16));
            }
            catch (Exception ex) { }
            i += 1;
        }

        base64 = Convert.ToBase64String(buffer);

        return base64;
    }

それが他の誰かに役立つことを願っています。

于 2013-12-05T16:23:51.517 に答える