convert hex string to base64

Question

Please help me on converting a hexstring to base64

here is the cede where I'm getting the exception

 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);

}

I'm getting an exception here::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)

looks like its duplicate of [this][1] [1]: stackoverflow.com/questions/7372268/… — banjara
– banjara, Commented Oct 13, 2012 at 6:55

aleroot · Accepted Answer · 2012-10-13 07:06:11Z

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;
          }

edited Oct 13, 2012 at 7:06

answered Oct 13, 2012 at 6:57

aleroot

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2 Comments

Sinha Over a year ago

Thanks aleroot!! that was very much helpful

Gichamba Over a year ago

@aleroot in your second example, shouldn't you be initializing i to 0 instead of 2?

cking24343 · Accepted Answer · 2013-12-05 16:23:51Z

-1

Found a similar solution, thought it might be good to share:

    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;
    }

Hope it helps someone else.

answered Dec 5, 2013 at 16:23

cking24343

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1 Comment

Matthew Over a year ago

good point about removing 0x as some firewalls will reject base65 containing 0x

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