我想设置一个特定的Drawable作为设备的壁纸,但所有的壁纸功能只接受位图。我不能使用WallpaperManager,因为我是pre 2.1。

另外,我的drawables是从网上下载的,并不存在于R.drawable中。


当前回答

方法1:你可以像这样直接转换成位图

Bitmap myLogo = BitmapFactory.decodeResource(context.getResources(), R.drawable.my_drawable);

方法2:你甚至可以把资源转换成可绘制的,从那里你可以得到像这样的位图

Bitmap myLogo = ((BitmapDrawable)getResources().getDrawable(R.drawable.logo)).getBitmap();

对于API > 22 getDrawable方法移动到ResourcesCompat类,因此你可以这样做

Bitmap myLogo = ((BitmapDrawable) ResourcesCompat.getDrawable(context.getResources(), R.drawable.logo, null)).getBitmap();

其他回答

非常简单的

Bitmap tempBMP = BitmapFactory.decodeResource(getResources(),R.drawable.image);

android-ktx有Drawable。toBitmap方法:https://android.github.io/android-ktx/core-ktx/androidx.graphics.drawable/android.graphics.drawable.-drawable/to-bitmap.html

从芬兰湾的科特林

val bitmap = myDrawable.toBitmap()

Bitmap Bitmap = BitmapFactory.decodeResource(context.getResources(), R.drawable.icon);

这将不会每次工作,例如,如果你的可绘制层列表可绘制,然后它给出一个空响应,所以作为一个替代方案,你需要绘制你的可绘制到画布,然后保存为位图,请参考下面的一杯代码。

public void drawableToBitMap(Context context, int drawable, int widthPixels, int heightPixels) {
    try {
        File file = new File(Environment.getExternalStoragePublicDirectory(Environment.DIRECTORY_DOWNLOADS) + "/", "drawable.png");
        FileOutputStream fOut = new FileOutputStream(file);
        Drawable drw = ResourcesCompat.getDrawable(context.getResources(), drawable, null);
        if (drw != null) {
            convertToBitmap(drw, widthPixels, heightPixels).compress(Bitmap.CompressFormat.PNG, 100, fOut);
        }
        fOut.flush();
        fOut.close();
    } catch (Exception e) {
        e.printStackTrace();
    }
}

private Bitmap convertToBitmap(Drawable drawable, int widthPixels, int heightPixels) {
    Bitmap bitmap = Bitmap.createBitmap(widthPixels, heightPixels, Bitmap.Config.ARGB_8888);
    Canvas canvas = new Canvas(bitmap);
    drawable.setBounds(0, 0, widthPixels, heightPixels);
    drawable.draw(canvas);
    return bitmap;
}

以上代码保存为drawable.png在下载目录

一个可绘制对象可以被绘制到画布上,一个画布可以被位图支持:

(更新到处理BitmapDrawables的快速转换,并确保创建的位图具有有效的大小)

public static Bitmap drawableToBitmap (Drawable drawable) {
    if (drawable instanceof BitmapDrawable) {
        return ((BitmapDrawable)drawable).getBitmap();
    }

    int width = drawable.getIntrinsicWidth();
    width = width > 0 ? width : 1;
    int height = drawable.getIntrinsicHeight();
    height = height > 0 ? height : 1;

    Bitmap bitmap = Bitmap.createBitmap(width, height, Bitmap.Config.ARGB_8888);
    Canvas canvas = new Canvas(bitmap); 
    drawable.setBounds(0, 0, canvas.getWidth(), canvas.getHeight());
    drawable.draw(canvas);

    return bitmap;
}

下面是@Chris提供的答案的Kotlin版本。Jenkins在这里:https://stackoverflow.com/a/27543712/1016462

fun Drawable.toBitmap(): Bitmap {
  if (this is BitmapDrawable) {
    return bitmap
  }

  val width = if (bounds.isEmpty) intrinsicWidth else bounds.width()
  val height = if (bounds.isEmpty) intrinsicHeight else bounds.height()

  return Bitmap.createBitmap(width.nonZero(), height.nonZero(), Bitmap.Config.ARGB_8888).also {
    val canvas = Canvas(it)
    setBounds(0, 0, canvas.width, canvas.height)
    draw(canvas)
  }
}

private fun Int.nonZero() = if (this <= 0) 1 else this