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java - Enforce 4 digit hex representation of a binary number

Below is a snippet of my java code.
//converts a binary string to hexadecimal
public static String binaryToHex (String binaryNumber)
{
BigInteger temp = new BigInteger(binaryNumber, 2);
return temp.toString(16).toUpperCase();
}
If I input "0000 1001 0101 0111" (without the spaces) as my String binaryNumber, the return value is 957. But ideally what I want is 0957 instead of just 957. How do I make sure to pad with zeroes if hex number is not 4 digits?
Thanks.

You do one of the following:
Manually pad with zeroes
Use String.format()
Manually pad with zeroes
Since you want extra leading zeroes when shorter than 4 digits, use this:
BigInteger temp = new BigInteger(binaryNumber, 2);
String hex = temp.toString(16).toUpperCase();
if (hex.length() < 4)
hex = "000".substring(hex.length() - 1) + hex;
return hex;
Use String.format()
BigInteger temp = new BigInteger(binaryNumber, 2);
return String.format("%04X", temp);
Note, if you're only expecting the value to be 4 hex digits long, then a regular int can hold the value. No need to use BigInteger.
In Java 8, do it by parsing the binary input as an unsigned number:
int temp = Integer.parseUnsignedInt(binaryNumber, 2);
return String.format("%04X", temp);

The BigInteger becomes an internal machine representation of whatever value you passed in as a String in binary format. Therefore, the machine does not know how many leading zeros you would like in the output format. Unfortunately, the method toString in BigInteger does not allow any kind of formatting, so you would have to do it manually if you attempt to use the code you showed.
I customized your code a bit to include leading zeros based on input string:
public static void main(String[] args) {
System.out.println(binaryToHex("0000100101010111"));
}
public static String binaryToHex(String binaryNumber) {
BigInteger temp = new BigInteger(binaryNumber, 2);
String hexStr = temp.toString(16).toUpperCase();
int b16inLen = binaryNumber.length()/4;
int b16outLen = hexStr.length();
int b16padding = b16inLen - b16outLen;
for (int i=0; i<b16padding; i++) {
hexStr=('0'+hexStr);
}
return hexStr;
}
Notice that the above solution counts up the base16 digits in the input and calculates the difference with the base16 digits in the output. So, it requires the user to input a full '0000' to be counted up. That is '000 1111' will be displayed as 'F' while '0000 1111' as '0F'.

Java decryption from byte to text

I want to write a java program which takes a text field with byte data. Output of my program should be string. How can I achieve that. Any inputs are appreciated.
Input is
85f960f0 82868260 f4f78486 60f8f6f
Output is string format like customer, hero, english..
I am planning to write a simple java program.
Thanks in advance.
Sorry for missing out details first time. I am in learning stages now.

Your question doesn't provide enough detail for a full answer. Assuming the fragment "85f960f0 82868260 f4f78486 60f8f6f" is the output you want...
convert a byte array to hexadecimal string using String.format() using the %x pattern within a loop.
Use %02x to pad each octet to 2 digits if necessary
if you need spaces every 8 characters you could do this by checking to see if the counter is divisible by 4 using the % operator.
For example.
byte[] valueFromTextField = "hello world foo bar".getBytes();
StringBuilder builder = new StringBuilder();
int i = 0;
for (byte element : valueFromTextField) {
if (i % 4 == 0 && builder.length() > 0) {
builder.append(" ");
}
builder.append(String.format("%02x", element));
i++;
}
System.out.println(builder.toString());
Output
68656c6c 6f20776f 726c6420 666f6f20 626172

Assuming you are having byte[] bytes and you can convert it using bytes.toString() OR you can change byte by byte
byte[] bites = new byte[]{24,4,72,56};
for(int i = 0; i < bites.length; i++)
System.out.println(new String(bites, i,1));

String hexadecimals = textField.getText();
hexadecimals = hexadecimals.replaceAll("[^0-9A-Fa-f]", ""); // Remove garbage
int nn = hexadecimals.length();
if (nn % 2 != 0) {
JOptionPane.showMessageDialog(null, "... must be even", JOptionPane.ERROR_MESSAGE);
return "";
}
byte[] bytes = new byte[nn / 2];
for (int i = 0; i < nn - 1; i += 2) {
int b = Integer.parseInt(hexadecimals.substring(i, i + 2), 16);
bytes[i] = (byte) b;
}
return new String(bytes, StandardCharsets.UTF_8);
This uses Integer.parseInt with base 16 (0-9A-F) in integer range to ignore negative byte values.
To convert those bytes to text (which in Java is Unicode to hold any combination of chars), one needs to know which text encoding those bytes are in. Here I use UTF-8, which however requires adherance to the UTF-8 multibyte format.

Split a string all alpha and all numeric

I have a List and I would like to split the first two characters (alpha characters) into a different string and then all the numbers that follow (they vary in length).
How could I do that?
String wholeString == "AB4578";
String alpha; // this has to be AB
String num; // this has to be 4578
Thank you very much in advance!

Tested and works:
String wholeString = "AB4578";
String alpha = null;
String num = null;
for (int i = 0; i < wholeString.length(); i++) {
if (wholeString.charAt(i) < 65) {
alpha = wholeString.substring(0, i);
num = wholeString.substring(i);
break;
}
}
With this approach both the A-z part and the 0-9 part can vary in size, it might not be very effective though considering it's calling charAt(...) for every char in the String.
Hope this helps.

String wholeString = "AB4578";
String alpha = wholeString.substring(0,2);
String num = wholeString.substring(2);
Must See
String.substring(int, int)

If the format is the same, then the answer is already provided. But if the format is not same than you can convert the string into char array and check each character against the ASCII values to check if it is an alphabet or a number.
char[] ch=wholestring.toCharArray();
Now you can apply a for loop for checking each character individually.
for(int l=0; l<ch.length;l++)
{
//code to check the characters
}
And you can separate both types in different strings using StringBuilder or forming two char arrays and then converting them to strings using
String.valueOf(chArray);
ASCII values - http://www.asciitable.com/

Try using the substring method for Strings.
Example:
String alpha = wholeString.substring(0,2);
String num = wholeString.substring(2);
http://download.oracle.com/javase/1.4.2/docs/api/java/lang/String.html#substring%28int%29

If the format is always the same you can just do this:
String wholeString = "AB4578";
String alpha = wholeString.substring(0, 2);
String num = wholeString.substring(2);

Recommend String API. You would need to use substring operations.

Get unicode value of a character

Is there any way in Java so that I can get Unicode equivalent of any character? e.g.
Suppose a method getUnicode(char c). A call getUnicode('÷') should return \u00f7.

You can do it for any Java char using the one liner here:
System.out.println( "\\u" + Integer.toHexString('÷' | 0x10000).substring(1) );
But it's only going to work for the Unicode characters up to Unicode 3.0, which is why I precised you could do it for any Java char.
Because Java was designed way before Unicode 3.1 came and hence Java's char primitive is inadequate to represent Unicode 3.1 and up: there's not a "one Unicode character to one Java char" mapping anymore (instead a monstrous hack is used).
So you really have to check your requirements here: do you need to support Java char or any possible Unicode character?

If you have Java 5, use char c = ...; String s = String.format ("\\u%04x", (int)c);
If your source isn't a Unicode character (char) but a String, you must use charAt(index) to get the Unicode character at position index.
Don't use codePointAt(index) because that will return 24bit values (full Unicode) which can't be represented with just 4 hex digits (it needs 6). See the docs for an explanation.
[EDIT] To make it clear: This answer doesn't use Unicode but the method which Java uses to represent Unicode characters (i.e. surrogate pairs) since char is 16bit and Unicode is 24bit. The question should be: "How can I convert char to a 4-digit hex number", since it's not (really) about Unicode.

private static String toUnicode(char ch) {
return String.format("\\u%04x", (int) ch);
}

char c = 'a';
String a = Integer.toHexString(c); // gives you---> a = "61"

I found this nice code on web.
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
public class Unicode {
public static void main(String[] args) {
System.out.println("Use CTRL+C to quite to program.");
// Create the reader for reading in the text typed in the console.
InputStreamReader inputStreamReader = new InputStreamReader(System.in);
BufferedReader bufferedReader = new BufferedReader(inputStreamReader);
try {
String line = null;
while ((line = bufferedReader.readLine()).length() > 0) {
for (int index = 0; index < line.length(); index++) {
// Convert the integer to a hexadecimal code.
String hexCode = Integer.toHexString(line.codePointAt(index)).toUpperCase();
// but the it must be a four number value.
String hexCodeWithAllLeadingZeros = "0000" + hexCode;
String hexCodeWithLeadingZeros = hexCodeWithAllLeadingZeros.substring(hexCodeWithAllLeadingZeros.length()-4);
System.out.println("\\u" + hexCodeWithLeadingZeros);
}
}
} catch (IOException ioException) {
ioException.printStackTrace();
}
}
}
Original Article

are you picky with using Unicode because with java its more simple if you write your program to use "dec" value or (HTML-Code) then you can simply cast data types between char and int
char a = 98;
char b = 'b';
char c = (char) (b+0002);
System.out.println(a);
System.out.println((int)b);
System.out.println((int)c);
System.out.println(c);
Gives this output
b
98
100
d

Convert a base25 String to binary String in Java

So I have a set of base digits like "BCDFGHJKLMNPQRSTVWXZ34679"
how do I convert a value say "D6CN96W6WT" to binary string in Java?

This should work (assuming 0,1 for you binary digits):
// your arbitrary digits
private static final String DIGITS = "BCDFGHJKLMNPQRSTVWXZ34679";
public String base25ToBinary(String base25Number) {
long value = 0;
char[] base25Digits = base25Number.toCharArray();
for (char digit : base25Digits) {
value = value * 25 + DIGITS.indexOf(digit);
}
return Long.toString(value, 2);
}

Off the top of my head, for base-25 strings.
Integer.toString(Integer.valueof(base25str, 25), 2)
Its a little unclear from your question whether you're talking about actual 0-9-Z bases, or a number encoding with an arbitrary list of symbols. I'm assuming the first, if its the later then you're out of luck on built-ins.