我要创建2 ArrayList中。一个拿着16种颜色，另外一个拿着139。

I want to create 2 ArrayList. One holding 16 colors, the other one holding 139.

我有一个颜色列表（包括RGB为255,126,32和十六进制为0xFFFF2552）。我想使用的ArrayList以后随机挑选颜色。

I have the list with colors (both RGB as 255,126,32 and Hex as 0xFFFF2552). I want to use the ArrayList to later pick random colors from.

我已经试过INT []，这是行不通的。我试过的ArrayList＆LT;整数GT; 和的ArrayList＆LT;颜色＆GT; 。我的问题是;我不明白如何将颜色添加到的ArrayList

I've tried int[], that doesn't work. I've tried ArrayList<Integer> and ArrayList<Color>. My problem is; I don't understand how to add the colors to the ArrayLists.

谢谢！

现在，我在探索这样的：

For now, I'm exploring this:

Color cBlue = new Color(0,0,255);
Color cRed = new Color(255,0,0);

ArrayList colors = new ArrayList();
colors.add(cBlue);
colors.add(cRed);

等等...

and so on...

我真的很喜欢 INT []颜色= =新的INT [] {4,5}; ，因为它只有一行code ...但如何我得到的色彩，到了后来，挑选？

I really like int[] colors = = new int[] {4,5}; because it's only one line of code... but how do I get colors in, to later on, pick from?

或..会是更好地存储在strings.xml中文件的颜色，然后从那里填写的ArrayList？如果是的话，我应该怎么办呢？

or.. would it be better to store the colors in a strings.xml file and then fill the ArrayList from there? If so, how should I do that?

谢谢！

推荐答案

您可以尝试：

int[] colors = new int[] {Color.rgb(1,1,1), Color.rgb(...)};

例如，但我不认为这是只用一条线的说法决定是个好主意。

For example, but I don't think it's a good idea to decide only using "one line" argument.

List<Integer> coloras = Arrays.asList(new Integer[]{Color.rgb(1, 1, 1), Color.rgb(...)});

也将工作。

您可以创建 arrays.xml 文件一个ArrayList

You can create an arraylist in arrays.xml file:

<resources>
    <string-array name="colors">        
        <item>#ff0000</item>
        <item>#00ff00</item>  
        <item>#0000ff</item>
    </string-array>
</resources>

然后使用循环读取它们：

Then use the loop to read them:

String[] colorsTxt = getApplicationContext().getResources().getStringArray(R.array.colors);
List<Integer> colors = new ArrayList<Integer>();
for (int i = 0; i < colorsTxt.length; i++) {
    int newColor = Color.parseColor(colorsTxt[i]);
    colors.add(newColor);
}

在我保持颜色列表中的意见是最方便易的解决方案。

In my opinion keeping colors in the list is the most convinient solution.

要采取从列表中选择颜色随机，你做的：

To take a color from the list randomly, you do:

int rand = new Random().nextInt(colors.size());
Integer color = colors.get(rand);