android从资产中获取位图或声音

Question

我需要从资产中获取位图和声音。 我尝试这样做：

BitmapFactory.decodeFile("file:///android_asset/Files/Numbers/l1.png");

像这样：

getBitmapFromAsset("Files/Numbers/l1.png");
    private Bitmap getBitmapFromAsset(String strName) {
        AssetManager assetManager = getAssets();
        InputStream istr = null;
        try {
            istr = assetManager.open(strName);
        } catch (IOException e) {
            e.printStackTrace();
        }
        Bitmap bitmap = BitmapFactory.decodeStream(istr);
        return bitmap;
    }

但我得到的只是免费空间，而不是图像。

这个怎么做？

Answer 1

public static Bitmap getBitmapFromAsset(Context context, String filePath) {
    AssetManager assetManager = context.getAssets();

    InputStream istr;
    Bitmap bitmap = null;
    try {
        istr = assetManager.open(filePath);
        bitmap = BitmapFactory.decodeStream(istr);
    } catch (IOException e) {
        // handle exception
    }

    return bitmap;
}

路径就是您的文件名 fx bitmap.png。 如果你使用子文件夹 bitmap/ 那么它的 bitmap/bitmap.png

Answer 2

使用此代码其工作

try {
    InputStream bitmap=getAssets().open("icon.png");
    Bitmap bit=BitmapFactory.decodeStream(bitmap);
    img.setImageBitmap(bit);
} catch (IOException e1) {
    // TODO Auto-generated catch block
    e1.printStackTrace();
}

更新

在解码 Bitmap 时，如果 Image size 非常大，我们更经常遇到内存溢出异常。 因此，阅读如何有效地显示图像的文章将对您有所帮助。

Answer 3

接受的答案永远不会关闭InputStream 。 这是在资产文件夹中获取Bitmap的实用方法：

/**
 * Retrieve a bitmap from assets.
 * 
 * @param mgr
 *            The {@link AssetManager} obtained via {@link Context#getAssets()}
 * @param path
 *            The path to the asset.
 * @return The {@link Bitmap} or {@code null} if we failed to decode the file.
 */
public static Bitmap getBitmapFromAsset(AssetManager mgr, String path) {
    InputStream is = null;
    Bitmap bitmap = null;
    try {
        is = mgr.open(path);
        bitmap = BitmapFactory.decodeStream(is);
    } catch (final IOException e) {
        bitmap = null;
    } finally {
        if (is != null) {
            try {
                is.close();
            } catch (IOException ignored) {
            }
        }
    }
    return bitmap;
}

Answer 4

Kotlin 方式，自动关闭输入流。 您必须自己处理异常。

private fun getBitmapFromAsset(context: Context, path: String): Bitmap =
        context.assets.open(path).use { BitmapFactory.decodeStream(it) }

Answer 5

获取存储在 Assets 文件夹中的图像位图的方法。

       public static Bitmap getBitmapFromAssets(Context context, String fileName, int width, int height) {
        AssetManager assetManager = context.getAssets();

        InputStream istr;
        Bitmap bitmap = null;
        try {
            final BitmapFactory.Options options = new BitmapFactory.Options();
            options.inJustDecodeBounds = true;

            istr = assetManager.open(fileName);

            // Calculate inSampleSize
            options.inSampleSize = calculateInSampleSize(options, width, height);

            // Decode bitmap with inSampleSize set
            options.inJustDecodeBounds = false;
            return BitmapFactory.decodeStream(istr, null, options);
        } catch (IOException e) {
            Log.e("hello", "Exception: " + e.getMessage());
        }

        return null;
    }

调整位图大小的方法。

 private static int calculateInSampleSize(
            BitmapFactory.Options options, int reqWidth, int reqHeight) {
        // Raw height and width of image
        final int height = options.outHeight;
        final int width = options.outWidth;
        int inSampleSize = 1;

        if (height > reqHeight || width > reqWidth) {

            final int halfHeight = height / 2;
            final int halfWidth = width / 2;

            // Calculate the largest inSampleSize value that is a power of 2 and keeps both
            // height and width larger than the requested height and width.
            while ((halfHeight / inSampleSize) >= reqHeight
                    && (halfWidth / inSampleSize) >= reqWidth) {
                inSampleSize *= 2;
            }
        }

        return inSampleSize;
    }

Answer 6

简短的 Kotlin 版本：

assets
   .open(name)
   .use(BitmapFactory::decodeStream)