[英]How to convert hex to rgb using Java?
How can I convert hex color to RGB code in Java?如何在 Java 中将十六进制颜色转换为 RGB 代码? Mostly in Google, samples are on how to convert from RGB to hex.大多数在 Google 中,示例是关于如何从 RGB 转换为十六进制的。
实际上,有一种更简单的(内置)方法可以做到这一点:
Color.decode("#FFCCEE");
I guess this should do it:我想这应该这样做:
/**
*
* @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 ) );
}
public static void main(String[] args) {
int hex = 0x123456;
int r = (hex & 0xFF0000) >> 16;
int g = (hex & 0xFF00) >> 8;
int b = (hex & 0xFF);
}
对于Android开发,我使用:
int color = Color.parseColor("#123456");
Here is a version that handles both RGB and RGBA versions:这是一个同时处理 RGB 和 RGBA 版本的版本:
/**
* 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;
}
you can do it simply as below:您可以简单地执行以下操作:
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例如
getRGB("444444") = 68,68,68
getRGB("FFFFFF") = 255,255,255
A hex color code is #RRGGBB十六进制颜色代码是#RRGGBB
RR, GG, BB are hex values ranging from 0-255 RR、GG、BB 是 0-255 范围内的十六进制值
Let's call RR XY where X and Y are hex character 0-9A-F, A=10, F=15让我们调用 RR XY 其中 X 和 Y 是十六进制字符 0-9A-F,A=10,F=15
The decimal value is X*16+Y十进制值为 X*16+Y
If RR = B7, the decimal for B is 11, so value is 11*16 + 7 = 183如果 RR = B7,则 B 的小数点为 11,因此值为 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;
}
For JavaFX对于JavaFX
import javafx.scene.paint.Color;
. .
Color whiteColor = Color.valueOf("#ffffff");
Lots of these solutions work, but this is an alternative.许多这些解决方案都有效,但这是另一种选择。
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.如果您不添加 4278190080 (#FF000000),则颜色的 Alpha 为 0 并且不会显示。
将其转换为整数,然后根据原始十六进制字符串的长度(分别为 3、6、9 或 12)将其 divmod 两次 16、256、4096 或 65536。
For Android Kotlin developers:对于 Android Kotlin开发人员:
"#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
}
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.要详细说明@xhh 提供的答案,您可以在返回之前附加红色、绿色和蓝色以将字符串格式化为“rgb(0,0,0)”。
/**
*
* @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();
}
If you don't want to use the AWT Color.decode, then just copy the contents of the method:如果您不想使用 AWT Color.decode,则只需复制该方法的内容:
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 Integer.decode 处理 # 或 0x,具体取决于字符串的格式
The easiest way:最简单的方法:
// 0000FF
public static Color hex2Rgb(String colorStr) {
return new Color(Integer.valueOf(colorStr, 16));
}
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.我认为这会奏效。
十六进制是基数 16,因此您可以使用 parseLong 使用基数 16 解析字符串:
Color newColor = new Color((int) Long.parseLong("FF7F0055", 16));
If you need to decode a HEXA string in following format #RRGGBBAA, you can use the following:如果需要解码以下格式#RRGGBBAA 的 HEXA 字符串,可以使用以下命令:
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:此外,如果要确保格式正确,可以使用此方法获得统一的结果:
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此方法会将所有可接受的格式(#RGB、#RGBA、#RRGGBB、RGB、RGBA、RRGGBB)转换为#RRGGBBAA
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]:前几天我一直在解决类似的问题,发现将十六进制颜色字符串转换为 int 数组 [alpha, r, g, b] 很方便:
/**
* 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;
}
Here is another faster version that handles RGBA versions:这是处理 RGBA 版本的另一个更快的版本:
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;
}
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);
Hexidecimal color codes are already rgb.十六进制颜色代码已经是 rgb。 The format is #RRGGBB格式为#RRGGBB
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