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如何获取资源文件夹中可绘制对象的路径名？我正在尝试在我的资源文件夹中获取图像并将其传递给函数以对其进行解码。

路径 "android.resource://com.myapp.example/"+ R.drawable.image" 不起作用。

 Bitmap bitmap;

 File image = new File("android.resource://com.myapp.example/"+ R.drawable.image);

 bitmap.decodeFile(image);


public Bitmap decodeFile(File f) {
    try {
        // decode image size
        BitmapFactory.Options o = new BitmapFactory.Options();
        o.inJustDecodeBounds = true;
        BitmapFactory.decodeStream(new FileInputStream(f), null, o);
        // Find the correct scale value. It should be the power of 2.
        final int REQUIRED_SIZE = 70;
        int width_tmp = o.outWidth, height_tmp = o.outHeight;
        int scale = 1;
        while (true) {
            if (width_tmp / 2 < REQUIRED_SIZE
                    || height_tmp / 2 < REQUIRED_SIZE)
                break;
            width_tmp /= 2;
            height_tmp /= 2;
            scale++;
        }

        // decode with inSampleSize
        BitmapFactory.Options o2 = new BitmapFactory.Options();
        o2.inSampleSize = scale;
        return BitmapFactory.decodeStream(new FileInputStream(f), null, o2);
    } catch (FileNotFoundException e) {
    }
    return null;
}
4

2 回答 2

6

BitmapFactory 具有解码资源的功能：

Bitmap BitmapFactory.decodeResource (Resources res, int id, BitmapFactory.Options opts)

例子：

bitmap.decodeFile(R.drawable.image);

public Bitmap decodeFile(int resId) {
    try {
        // decode image size
        BitmapFactory.Options o = new BitmapFactory.Options();
        o.inJustDecodeBounds = true;
        BitmapFactory.decodeResource(getContext().getResources(), resId, o);
        // Find the correct scale value. It should be the power of 2.
        final int REQUIRED_SIZE = 70;
        int width_tmp = o.outWidth, height_tmp = o.outHeight;
        int scale = 1;
        while (true) {
            if (width_tmp / 2 < REQUIRED_SIZE
                    || height_tmp / 2 < REQUIRED_SIZE)
                break;
            width_tmp /= 2;
            height_tmp /= 2;
            scale++;
        }

        // decode with inSampleSize
        BitmapFactory.Options o2 = new BitmapFactory.Options();
        o2.inSampleSize = scale;
        return BitmapFactory.decodeResource(getContext().getResources(), resId, o2);
    } catch (Exception e) {
    }
    return null;
}
于 2013-09-18T18:13:33.587 回答
0

不，你无法获取路径，因为资源文件夹被编译成apk，但你可以像这样抓取资源：

Resources rsrc= getResources();
Drawable drawable= rsrc.getDrawable(R.drawable.datImage);
Bitmap bm = BitmapFactory.decodeResource(rsrc,drawable, opts);
于 2013-09-18T18:41:39.643 回答