I\'m trying to encrypt some integers in java using java.security and javax.crypto.
The problem seems to be that the Cipher class only encrypts byte arrays. I can\'t d
create a 4-byte array and copy the int to the array in 4 steps, with bitwise ANDs and bitshifting, like Paulo said.
But remember that block algorithms such as AES and DES work with 8 or 16 byte blocks so you will need to pad the array to what the algorithm needs. Maybe leave the first 4 bytes of an 8-byte array as 0's, and the other 4 bytes contain the integer.
You can turn ints into a byte[] using a DataOutputStream, like this:
ByteArrayOutputStream baos = new ByteArrayOutputStream ();
DataOutputStream dos = new DataOutputStream (baos);
dos.writeInt (i);
byte[] data = baos.toByteArray();
// do encryption
Then to decrypt it later:
byte[] decrypted = decrypt (data);
ByteArrayInputStream bais = new ByteArrayInputStream (data);
DataInputStream dis = new DataInputStream (bais);
int j = dis.readInt();
Just use:
Integer.toString(int).getBytes();
Make sure you use your original int and getBytes() will return a byte array. No need to do anything else complicated.
To convert back:
Integer.parseInt(encryptedString);
My Simple Solution is that Encrypt Integer to the String by shifting ASCII Value of the Integer by the secret key you Provide.
Here is the Solution:
public String encodeDiscussionId(int Id) {
String tempEn = Id + "";
String encryptNum ="";
for(int i=0;i<tempEn.length();i++) {
int a = (int)tempEn.charAt(i);
a+=148113;
encryptNum +=(char)a;
}
return encryptNum;
}
public Integer decodeDiscussionId(String encryptText) {
String decodeText = "";
for(int i=0;i<encryptText.length();i++) {
int a= (int)encryptText.charAt(i);
a -= 148113;
decodeText +=(char)a;
}
int decodeId = Integer.parseInt(decodeText);
return decodeId;
}
Steps to Encode:
String temp = givenInt + ""
Integer
into Character and concatenate to the String encryptNum
and finally return it.Steps to Decode:
decodeText
. As previous encode output is always String '???'
and vary according to number of digits of input Id
.
Just use NIO. It's designed for this specific purpose. ByteBuffer and IntBuffer will do what you need quickly, efficiently, and elegantly. It'll handle big/little endian conversion, "direct" buffers for high performance IO, and you can even mix data types into the byte buffer.
Convert integers into bytes:
ByteBuffer bbuffer = ByteBuffer.allocate(4*theIntArray.length);
IntBuffer ibuffer = bbuffer.asIntBuffer(); //wrapper--doesn't allocate more memory
ibuffer.put(theIntArray); //add your int's here; can use
//array if you want
byte[] rawBytes = bbuffer.array(); //returns array backed by bbuffer--
//i.e. *doesn't* allocate more memory
Convert bytes into integers:
ByteBuffer bbuffer = ByteBuffer.wrap(rawBytes);
IntBuffer ibuffer = bbuffer.asIntBuffer();
while(ibuffer.hasRemaining())
System.out.println(ibuffer.get()); //also has bulk operators
I have found the following code that may help you, since Integer in Java is always 4 bytes long.
public static byte[] intToFourBytes(int i, boolean bigEndian) {
if (bigEndian) {
byte[] data = new byte[4];
data[3] = (byte) (i & 0xFF);
data[2] = (byte) ((i >> 8) & 0xFF);
data[1] = (byte) ((i >> 16) & 0xFF);
data[0] = (byte) ((i >> 24) & 0xFF);
return data;
} else {
byte[] data = new byte[4];
data[0] = (byte) (i & 0xFF);
data[1] = (byte) ((i >> 8) & 0xFF);
data[2] = (byte) ((i >> 16) & 0xFF);
data[3] = (byte) ((i >> 24) & 0xFF);
return data;
}
}
You can find more information about the bigEndian parameter here: http://en.wikipedia.org/wiki/Endianness