创建一个新的可绘制颜色

我试图转换一个十六进制值为整型，以便我可以创建一个新的颜色绘制。我不确定这是否可行，但根据 文件，应该可行。它明显要求

Public ColorDrawable (int color)

在 API 级别1中添加创建具有指定的 颜色。

参数 颜色要绘制的颜色。

因此，我的代码无法工作，因为我得到了一个无效的 int: “ FF6666”错误。有什么办法吗？

int decode = Integer.decode("FF6666");
ColorDrawable colorDrawable = new ColorDrawable(decode);
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最佳答案

Since you're talking about hex you have to start with 0x and don't forget the opacity.

So basically: 0xFFFF6666

ColorDrawable cd = new ColorDrawable(0xFFFF6666);

You can also create a new colors.xml file into /res and define the colors like:

<?xml version="1.0" encoding="utf-8"?>
<resources>
<color name="mycolor">#FF6666</color>
</resources>

and simply get the color defined in R.color.mycolor

getResources().getColor(R.color.mycolor)
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I think you have to use :

public static int parseColor (String colorString)

Added in API level 1 Parse the color string, and return the corresponding color-int. If the string cannot be parsed, throws an IllegalArgumentException exception. Supported formats are: #RRGGBB #AARRGGBB red, blue, green, black, white, gray, cyan, magenta, yellow, lightgray, darkgray, grey, lightgrey, darkgrey, aqua, fuschia, lime, maroon, navy, olive, purple, silver, teal

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It should be like this...

ColorDrawable cd = new ColorDrawable(0xffff6666);

Note I used 8 hex digits, not 6 hex digit . which add to transparency

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For using with ContextCompat and rehuse the color you can do something like this:

ColorDrawable colorDrawable = new ColorDrawable(ContextCompat.getColor(this, R.color.white));
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By followingthe above advice,to be a summary of this question:

  1. ColorDrawable colorDrawable = new ColorDrawable(Color.parseColor("#ce9b2c"));`

  2. ColorDrawable colorDrawable = new ColorDrawable(0xFFCE9B2C); Note there is 8 hex digits, not 6 hex digit,which no work. Case all

  3. ColorDrawable colorDrawable = new ColorDrawable(ContextCompat.getColor(mContext,R.color.default_color));

Selecting up to you!

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This is how I converted a Hex color to int and applied to a Background of a View

Let's say that we have a color #8080000.

1) Hex to int conversion

int myColor = Color.parseColor("#808000");

2) Set background

view.setBackgroundColor(context.getColor(myColor));
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Xamarin/Maui :

  protected override void OnCreate(Bundle savedInstanceState)
{
base.OnCreate(savedInstanceState);


Window.SetBackgroundDrawable(new ColorDrawable(ColorExtensions.ToAndroid(Colors.Black)));
}

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