Question

我在这里有一个FileOutputstream：

    public void SaveImage(Bitmap default_b) {

    String root = Environment.getExternalStorageDirectory().toString();
    File myDir = new File(root + "/saved_images");
    myDir.mkdirs();
    Random generator = new Random();
    int n = 100000;
    n = generator.nextInt(n);
    String fname = "Image-" + n +".jpg";
    File file = new File (myDir, fname);
    if (file.exists()) file.delete();
    try {
        FileOutputStream out = openFileOutput("file", Context.MODE_WORLD_WRITEABLE);
        default_b.compress(Bitmap.CompressFormat.JPEG, 90, out);
        out.flush();
        out.close();
    } catch (Exception e) {
        e.printStackTrace();
    }
}

但我在这一行上收到错误：

FileOutputStream out = openFileOutput("file", Context.MODE_WORLD_WRITEABLE);

说openFileOutput是未定义的*（请注意，这是在基本适配器类中，当我尝试从上下文中执行此操作时，Context.getApplicationContext().openFileOutput会收到错误“无法对非静态方法getApplicationContext（）来自类型Context“）我还得到了不推荐使用Context.MODE_WORLD_WRITABLE的警告。

然后我在这里有我的FileInputstream（在另一个扩展BaseAdapter的类中）：

@Override
public View getView(int position, View convertView, ViewGroup parent) {
    // Try to reuse the views
    ImageView view = (ImageView) convertView;
    // if convert view is null then create a new instance else reuse it
    if (view == null) {
        view = new ImageView(mContextGV);
        Log.d("GridViewAdapter", "new imageView added");
    }
    try {
        FileInputStream in = new openFileInput("file");
        BufferedInputStream buf = new BufferedInputStream(in);
        byte[] bitMapA = new byte[buf.available()];
        buf.read(bitMapA);
        Bitmap bM = BitmapFactory.decodeByteArray(bitMapA, 0, bitMapA.length);
        view.setImageBitmap(bM);
        if (in != null) {
            in.close();
        }
        if (buf != null) {
            buf.close();
        }
    } catch (Exception e) {
        e.printStackTrace();
    }
    view.setImageResource(drawables.get(position));
    view.setScaleType(ImageView.ScaleType.CENTER_CROP);
    view.setLayoutParams(new android.widget.GridView.LayoutParams(70, 70));
    view.setTag(String.valueOf(position));
    return view;
}

这一行有错误：

FileInputStream in = new openFileInput("file");

说openFileInput是未定义的，当我将其更改为：

时再次

FileInputStream in = new Context.getApplicationContext().openFileInput("file");

我收到一条错误消息，说“Context.getApplicationContext（）无法解析为类型”

Answer 1

尝试替换文件OutputStream代码

String fname = "Image-" + n ;
File file = new File (myDir, fname);
if (file.exists()) file.delete();
try {
   File f = new File(Environment.getExternalStorageDirectory().getAbsolutePath()
      + "/" + fname +".png");
FileOutputStream fos = new  FileOutputStream(f);
default_b.compress(Bitmap.CompressFormat.PNG,90,fos);
} catch (Exception e) {
    e.printStackTrace();
}

它的作品非常适合我保存图像

FileOutputstream和FileInputStream出错

1 个答案: