如何通过使用十六进制codeS检测所有色调？ [英] How to detect all color tones by using hex codes?

问题描述

我想开发一个Android应用程序将采取一个物体的图片，然后检测物体的颜色。我想说明的颜色有对象的用户。我根据密度和亮度的答案的帮助下，我的问题在这个环节中实现的检测颜色：

I'm trying to develop an Android application which will take the picture of an object and then detect the color of that object. I want to show user which color has the object. I've implemented the detecting color according to density and luminance with the help of answers to my question in this link:

<一个href=\"http://stackoverflow.com/questions/28670901/what-is-the-best-way-to-implement-color-detection-in-android/28673282?noredirect=1#comment45727933_28673282\">What是落实颜色检测的Andr​​oid的最佳途径？

在这一点上，我能够得到的颜色为十六进制code。我真正想要做的是，能够告知用户哪些颜色是十六进制code。

At this point, I'm able to get color as a hex code. What I really want to do is that being able to inform user about which color is that hex code.

我不希望只是限制我的应用程序，以检测为主色调，所以我想它来检测许多不同的颜色。

I don't want to limit my application just to detect main colors so I want it to detect many different colors.

我如何使用这些十六进制$ C $这样做CS？

How can I do this by using these hex codes?

感谢您提前。

推荐答案

很多的努力，找出我的问题的有效解决方案，终于我知道了！在这里，下面的人谁将会解决这个问题面临着能够找到一个替代的解决方案：

After a lot of effort to find out an efficient solution for my problem, finally I got it! Here below anyone who will face with this problem can find an alternative solution:

生成找到最接近的颜色名称这是你的数据库的方法：

Generate a method to find closest color name which is in your database:

private String findClosedColor(String hexColor) {
    int rgb[] = hexToRGB(hexColor);
    int min = 3 * (int) pow(256, 2) + 1;
    ArrayList<HashMap<String, String>> colorList = getColorList();

    String colorName = null;
    int i;
    int len = colorList.size();
    for (i = 0; i < len; i++) {
        HashMap<String, String> map = colorList.get(i);
        String colorCode = map.get("code");
        Log.w("myApp", "HashMap'ten gelen colorCode:" + colorCode);
        if (colorCode != null) {
            int df = rgbDistance(hexToRGB(colorCode), rgb);
            if (df < min) {
                min = df;
                colorName = map.get("name");
            }
        }
    }
    return colorName;
}

private int rgbDistance(int[] c1, int[] c2) {
    return ( (int) pow(c1[0] - c2[0], 2)) + ((int) pow(c1[1] - c2[1], 2)) + ((int) pow(c1[2] - c2[2], 2));
}

private int[] hexToRGB( String hexCode)
{
    int returnValue[] = new int[3];

    if (hexCode.charAt(0) == '#')
    {
        hexCode = hexCode.substring(1);
    }

    if (hexCode.length() < 6)
    {
        returnValue[0] = -1;
        returnValue[1] = -1;
        returnValue[2] = -1;
    }
    else
    {
        int r = fromHex(hexCode.substring(0, 2));
        int g = fromHex(hexCode.substring(2, 4));
        int b = fromHex(hexCode.substring(4, 6));

        returnValue[0] = r;
        returnValue[1] = g;
        returnValue[2] = b;

    }
    return returnValue;
}

private int fromHex( String n) {
    n = n.toUpperCase();
    if (n.length() < 2)
        return -1;
    int f1 = letterToCode(n.charAt(0));
    int f2 = letterToCode(n.charAt(1));
    if (f1 == -1 || f2 == -1) {
        return -1;
    } else {
        return f1 * 16 + f2;
    }
}

private int letterToCode(char n) {
    switch (n) {
        case '0': return 0;
        case '1': return 1;
        case '2': return 2;
        case '3': return 3;
        case '4': return 4;
        case '5': return 5;
        case '6': return 6;
        case '7': return 7;
        case '8': return 8;
        case '9': return 9;
        case 'A': return 10;
        case 'B': return 11;
        case 'C': return 12;
        case 'D': return 13;
        case 'E': return 14;
        case 'F': return 15;
        default: return -1;
    }
}

getColorList（）函数返回从我的数据库中的颜色列表。有了这个解决方案，我可以很容易地在我的数据库选择接近检测名每六角code。

getColorList() function returns the color list from my database. With this solution, I can easily detect every hex code by choosing closer name in my database.

最好的问候给大家......

Best Regards to everyone...

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