How to convert hex to rgb using Java?

Question

How can I convert hex color to RGB code in Java? Mostly in Google, samples are on how to convert from RGB to hex.

Answer 1

Actually, there's an easier (built in) way of doing this:

Color.decode("#FFCCEE");

Answer 2

I guess this should do it:

/**
 * 
 * @param colorStr e.g. "#FFFFFF"
 * @return 
 */
public static Color hex2Rgb(String colorStr) {
    return new Color(
            Integer.valueOf( colorStr.substring( 1, 3 ), 16 ),
            Integer.valueOf( colorStr.substring( 3, 5 ), 16 ),
            Integer.valueOf( colorStr.substring( 5, 7 ), 16 ) );
}

Answer 3

public static void main(String[] args) {
    int hex = 0x123456;
    int r = (hex & 0xFF0000) >> 16;
    int g = (hex & 0xFF00) >> 8;
    int b = (hex & 0xFF);
}

Answer 4

For Android development, I use:

int color = Color.parseColor("#123456");

Answer 5

Here is a version that handles both RGB and RGBA versions:

/**
 * Converts a hex string to a color. If it can't be converted null is returned.
 * @param hex (i.e. #CCCCCCFF or CCCCCC)
 * @return Color
 */
public static Color HexToColor(String hex) 
{
    hex = hex.replace("#", "");
    switch (hex.length()) {
        case 6:
            return new Color(
            Integer.valueOf(hex.substring(0, 2), 16),
            Integer.valueOf(hex.substring(2, 4), 16),
            Integer.valueOf(hex.substring(4, 6), 16));
        case 8:
            return new Color(
            Integer.valueOf(hex.substring(0, 2), 16),
            Integer.valueOf(hex.substring(2, 4), 16),
            Integer.valueOf(hex.substring(4, 6), 16),
            Integer.valueOf(hex.substring(6, 8), 16));
    }
    return null;
}

Answer 6

you can do it simply as below:

 public static int[] getRGB(final String rgb)
{
    final int[] ret = new int[3];
    for (int i = 0; i < 3; i++)
    {
        ret[i] = Integer.parseInt(rgb.substring(i * 2, i * 2 + 2), 16);
    }
    return ret;
}

For Example

getRGB("444444") = 68,68,68   
getRGB("FFFFFF") = 255,255,255

Answer 7

A hex color code is #RRGGBB

RR, GG, BB are hex values ranging from 0-255

Let's call RR XY where X and Y are hex character 0-9A-F, A=10, F=15

The decimal value is X*16+Y

If RR = B7, the decimal for B is 11, so value is 11*16 + 7 = 183

public int[] getRGB(String rgb){
    int[] ret = new int[3];
    for(int i=0; i<3; i++){
        ret[i] = hexToInt(rgb.charAt(i*2), rgb.charAt(i*2+1));
    }
    return ret;
}

public int hexToInt(char a, char b){
    int x = a < 65 ? a-48 : a-55;
    int y = b < 65 ? b-48 : b-55;
    return x*16+y;
}

Answer 8

For JavaFX

import javafx.scene.paint.Color;

.

Color whiteColor = Color.valueOf("#ffffff");

Answer 9

Lots of these solutions work, but this is an alternative.

String hex="#00FF00"; // green
long thisCol=Long.decode(hex)+4278190080L;
int useColour=(int)thisCol;

If you don't add 4278190080 (#FF000000) the colour has an Alpha of 0 and won't show.

Answer 10

Convert it to an integer, then divmod it twice by 16, 256, 4096, or 65536 depending on the length of the original hex string (3, 6, 9, or 12 respectively).

Answer 11

For Android Kotlin developers:

"#FFF".longARGB()?.let{ Color.parceColor(it) }
"#FFFF".longARGB()?.let{ Color.parceColor(it) }

fun String?.longARGB(): String? {
    if (this == null || !startsWith("#")) return null
    
//    #RRGGBB or #AARRGGBB
    if (length == 7 || length == 9) return this

//    #RGB or #ARGB
    if (length in 4..5) {
        val rgb = "#${this[1]}${this[1]}${this[2]}${this[2]}${this[3]}${this[3]}"
        if (length == 5) {
            return "$rgb${this[4]}${this[4]}"
        }
        return rgb
    }

    return null
}

Answer 12

To elaborate on the answer @xhh provided, you can append the red, green, and blue to format your string as "rgb(0,0,0)" before returning it.

/**
* 
* @param colorStr e.g. "#FFFFFF"
* @return String - formatted "rgb(0,0,0)"
*/
public static String hex2Rgb(String colorStr) {
    Color c = new Color(
        Integer.valueOf(hexString.substring(1, 3), 16), 
        Integer.valueOf(hexString.substring(3, 5), 16), 
        Integer.valueOf(hexString.substring(5, 7), 16));

    StringBuffer sb = new StringBuffer();
    sb.append("rgb(");
    sb.append(c.getRed());
    sb.append(",");
    sb.append(c.getGreen());
    sb.append(",");
    sb.append(c.getBlue());
    sb.append(")");
    return sb.toString();
}

Answer 13

If you don't want to use the AWT Color.decode, then just copy the contents of the method:

int i = Integer.decode("#FFFFFF");
int[] rgb = new int[]{(i >> 16) & 0xFF, (i >> 8) & 0xFF, i & 0xFF};

Integer.decode handles the # or 0x, depending on how your string is formatted

Answer 14

The easiest way:

// 0000FF
public static Color hex2Rgb(String colorStr) {
    return new Color(Integer.valueOf(colorStr, 16));
}

Answer 15

public static Color hex2Rgb(String colorStr) {
    try {
        // Create the color
        return new Color(
                // Using Integer.parseInt() with a radix of 16
                // on string elements of 2 characters. Example: "FF 05 E5"
                Integer.parseInt(colorStr.substring(0, 2), 16),
                Integer.parseInt(colorStr.substring(2, 4), 16),
                Integer.parseInt(colorStr.substring(4, 6), 16));
    } catch (StringIndexOutOfBoundsException e){
        // If a string with a length smaller than 6 is inputted
        return new Color(0,0,0);
    }
}

public static String rgbToHex(Color color) {
    //      Integer.toHexString(), built in Java method        Use this to add a second 0 if the
    //     .Get the different RGB values and convert them.     output will only be one character.
    return Integer.toHexString(color.getRed()).toUpperCase() + (color.getRed() < 16 ? 0 : "") + // Add String
            Integer.toHexString(color.getGreen()).toUpperCase() + (color.getGreen() < 16 ? 0 : "") +
            Integer.toHexString(color.getBlue()).toUpperCase() + (color.getBlue() < 16 ? 0 : "");
}

I think that this wil work.

Answer 16

Hex is base 16, so you can parse the string with parseLong using a radix of 16 :

Color newColor = new Color((int) Long.parseLong("FF7F0055", 16));

Answer 17

If you need to decode a HEXA string in following format #RRGGBBAA, you can use the following:

private static Color convert(String hexa) {
  var value = Long.decode(hexa);
  return new Color(
    (int) (value >> 24) & 0xFF,
    (int) (value >> 16) & 0xFF,
    (int) (value >> 8) & 0xFF,
    (int) (value & 0xFF)
  );
}

Furthermore, if you want to ensure correct format, you can use this method to get a uniform result:

private static String format(String raw) {
  var builder = new StringBuilder(raw);
  if (builder.charAt(0) != '#') {
    builder.insert(0, '#');
  }
  if (builder.length() == 9) {
    return builder.toString();
  } else if (builder.length() == 7) {
    return builder.append("ff").toString();
  } else if (builder.length() == 4) {
    builder.insert(builder.length(), 'f');
  } else if (builder.length() != 5) {
    throw new IllegalStateException("unsupported format");
  }
  for (int index = 1; index <= 7; index += 2) {
    builder.insert(index, builder.charAt(index));
  }
  return builder.toString();
}

This method will turn every accepted format (#RGB, #RGBA, #RRGGBB, RGB, RGBA, RRGGBB) into #RRGGBBAA

Answer 18

The other day I'd been solving the similar issue and found convenient to convert hex color string to int array [alpha, r, g, b]:

 /**
 * Hex color string to int[] array converter
 *
 * @param hexARGB should be color hex string: #AARRGGBB or #RRGGBB
 * @return int[] array: [alpha, r, g, b]
 * @throws IllegalArgumentException
 */

public static int[] hexStringToARGB(String hexARGB) throws IllegalArgumentException {

    if (!hexARGB.startsWith("#") || !(hexARGB.length() == 7 || hexARGB.length() == 9)) {

        throw new IllegalArgumentException("Hex color string is incorrect!");
    }

    int[] intARGB = new int[4];

    if (hexARGB.length() == 9) {
        intARGB[0] = Integer.valueOf(hexARGB.substring(1, 3), 16); // alpha
        intARGB[1] = Integer.valueOf(hexARGB.substring(3, 5), 16); // red
        intARGB[2] = Integer.valueOf(hexARGB.substring(5, 7), 16); // green
        intARGB[3] = Integer.valueOf(hexARGB.substring(7), 16); // blue
    } else hexStringToARGB("#FF" + hexARGB.substring(1));

    return intARGB;
}

Answer 19

Here is another faster version that handles RGBA versions:

public static int hexToIntColor(String hex){
    int Alpha = Integer.valueOf(hex.substring(0, 2), 16);
    int Red = Integer.valueOf(hex.substring(2, 4), 16);
    int Green = Integer.valueOf(hex.substring(4, 6), 16);
    int Blue = Integer.valueOf(hex.substring(6, 8), 16);
    Alpha = (Alpha << 24) & 0xFF000000;
    Red = (Red << 16) & 0x00FF0000;
    Green = (Green << 8) & 0x0000FF00;
    Blue = Blue & 0x000000FF;
    return Alpha | Red | Green | Blue;
}

Answer 20

For shortened hex code like #fff or #000

int red = "colorString".charAt(1) == '0' ? 0 : 
     "colorString".charAt(1) == 'f' ? 255 : 228;  
int green =
     "colorString".charAt(2) == '0' ? 0 :  "colorString".charAt(2) == 'f' ?
     255 : 228;  
int blue = "colorString".charAt(3) == '0' ? 0 : 
     "colorString".charAt(3) == 'f' ? 255 : 228;

Color.rgb(red, green,blue);

Answer 21

Hexidecimal color codes are already rgb. The format is #RRGGBB