java - 需要使十六进制 0xff****** 字符串更暗（* = 任何十六进制字符）

Question

我正在尝试获取用户输入的十六进制整数并使其变暗。有任何想法吗？

Answer 1

只需进行二进制减法：

int red = 0xff0000;
int darkerRed = (0xff0000 - 0x110000);
int programmaticallyDarkerRed;

您甚至可以使用循环使其更暗：

for(int i = 1; i < 15; i++)
{
    programmaticallyDarkerRed = (0xff0000 - (i * 0x110000));
}

它越接近0x000000或变黑，它就会越暗。

Answer 2

我会用Color.darker.

Color c = Color.decode(hex).darker();

Answer 3

您可以将十六进制值转换为 aColor然后使Color对象变暗

// Convert the hex to an color (or use what ever method you want)
Color color = Color.decode(hex);

// The fraction of darkness you want to apply
float fraction = 0.1f;

// Break the color up
int red = color.getRed();
int blue = color.getBlue();
int green = color.getGreen();
int alpha = color.getAlpha();

// Convert to hsb
float[] hsb = Color.RGBtoHSB(red, green, blue, null);
// Decrease the brightness
hsb[2] = Math.min(1f, hsb[2] * (1f - fraction));
// Re-assemble the color
Color hSBColor = Color.getHSBColor(hsb[0], hsb[1], hsb[2]);

// If you need it, you will need to reapply the alpha your self

更新

要将其恢复为十六进制，您可以尝试类似

int r = color.getRed();
int g = color.getGreen();
int b = color.getBlue();

String rHex = Integer.toString(r, 16);
String gHex = Integer.toString(g, 16);
String bHex = Integer.toString(b, 16);

String hexValue = (rHex.length() == 2 ? "" + rHex : "0" + rHex)
                + (gHex.length() == 2 ? "" + gHex : "0" + gHex)
                + (bHex.length() == 2 ? "" + bHex : "0" + bHex);

int intValue = Integer.parseInt(hex, 16); 

现在，如果那不太正确，我会看看是否有任何 SO 或 Google 的答案