需要使一个十六进制0xff ******字符串较深（* =任何十六进制字符）

Question

我想要一个用户输入十六进制int并使其更暗。有任何想法吗？

Answer 1

只是做二进制减法：

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

你甚至可以使其与一个循环较暗：

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

它得到0x000000或黑色越近，越暗它将会。

Answer 2

您可以在十六进制值转换为Color，然后变暗Color对象

// 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 

UPDATE

拿回来为十六进制，你可以尝试像

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，如果这不是很对，我会看看，看看是否有任何答案的S O或谷歌

Answer 3

我会用Color.darker。

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