Question

我正在尝试创建一个应用，用户可以从图库中选择他们的个人资料图片。我决定将他们的个人资料图片保存为我的数据库Blob。我能够保存图像甚至检索它。问题是，我无法替换它，或者每当我再次点击它时，应用程序停止工作，当我检查我存储图像的表格时，它说“返回了太多数据......”

public class AccountFragment extends Fragment implements OnClickListener {
private LoginDataBaseAdapter loginDataBaseAdapter;
Bitmap image;
Bitmap bitmap;
String picture_location;
TextView textTargetUri;
ImageView targetImage;
public static final String MyPREFERENCES = "MyPrefs" ;

    @Override
    public View onCreateView(LayoutInflater inflater, ViewGroup container,
            Bundle savedInstanceState) {
         // create a instance of SQLite Database
        loginDataBaseAdapter = new LoginDataBaseAdapter(getActivity());
         loginDataBaseAdapter=loginDataBaseAdapter.open();

      //intialize variables

            textTargetUri = (TextView) rootView.findViewById(R.id.targeturi);

            targetImage=(ImageView) rootView.findViewById(R.id.profpic);

            targetImage.setOnClickListener(new ImageView.OnClickListener(){
            @Override
            public void onClick(View arg0) {
                Intent   intent = new Intent(Intent.ACTION_PICK,
                android.provider.MediaStore.Images.Media.EXTERNAL_CONTENT_URI);
                startActivityForResult(intent, 0);
            }});



            showpic();


        return rootView;


         }

    @Override
    public void onActivityResult( int requestCode, int resultCode, Intent data) {
        super.onActivityResult(requestCode, resultCode, data);

        if (resultCode == Activity.RESULT_OK){
            Uri targetUri = data.getData();
            picture_location = targetUri.toString();
            textTargetUri.setText(targetUri.toString());
            Bitmap bitmap;
            try {

                bitmap = BitmapFactory.decodeStream(getActivity().getContentResolver().openInputStream(targetUri));  
                ByteArrayOutputStream stream = new ByteArrayOutputStream();
                bitmap.compress(Bitmap.CompressFormat.JPEG, 100, stream);
                byte[] byteArray = stream.toByteArray();
                loginDataBaseAdapter.insertPhoto(byteArray);
                showpic();


            } 
            catch (FileNotFoundException e){

                e.printStackTrace();
            }

        }   
    }

    @Override
    public void onClick(View arg0) {
        // TODO Auto-generated method stub

    }

    public void showpic() {



         Cursor cursor = loginDataBaseAdapter.fetchProfileImageFromDatabase();  

         if(cursor != null)  
         {  
             cursor.moveToFirst();  


                    byte[] data = cursor.getBlob(cursor.getColumnIndex("Path"));  
                    ByteArrayInputStream imageStream = new ByteArrayInputStream(data);  
                    Bitmap theImage = BitmapFactory.decodeStream(imageStream);  
                    targetImage.setImageBitmap(theImage);  

                }  




             cursor.close();  
        }  
    }

和我的数据库处理程序：

     //IMAGE
            public static final String Profpic_TABLE = "ProfilePic";
            public static final String KEY_ProfpicID = "_id";
            public static final String KEY_ProfPic = "Path";


     //ProfilePic-Table 
            static final String DATABASE_ProfPic =
                    "create table " + Profpic_TABLE + " ("
                    + KEY_ProfpicID + " integer primary key DEFAULT 1, "
                    + KEY_ProfPic + " BLOB);";  


        public long insertPhoto(byte[] EImage) {

        db.execSQL("delete from "+ Profpic_TABLE);

        try {
            System.out.println("Function call : ");
            ContentValues values = new ContentValues();

            values.put(KEY_ProfPic, EImage);
            return db.insert(Profpic_TABLE, null, values);


        } catch (Exception e) {
            e.printStackTrace();
            return 0;
        }
    }

      public Cursor fetchProfileImageFromDatabase()  
        {  
            return  db.rawQuery("SELECT Path FROM ProfilePic where _id = 1 " , null);  
        } 

    }

Answer 1

我终于能够解决它了。事实证明我必须这样做：

        public void showpic() {

        LoginDataBaseAdapter db = loginDataBaseAdapter.open();
         boolean emptytab = false;
         boolean empty = db.checkPic(null, emptytab);

        //Cursor cursor = loginDataBaseAdapter.fetchProfileImageFromDatabase();  

         if(empty==false)  
         {  
             Cursor cursor = loginDataBaseAdapter.fetchProfileImageFromDatabase();
             cursor.moveToFirst();
                   byte[] data = cursor.getBlob(cursor.getColumnIndex("Path"));  
                    ByteArrayInputStream imageStream = new ByteArrayInputStream(data);  
                    Bitmap theImage = BitmapFactory.decodeStream(imageStream);  
                    targetImage.setImageBitmap(theImage);  
            cursor.close(); 
                }  





        }

在我的db适配器上我添加了这个：

public boolean checkPic（String count，Object object）{

        boolean empty = false;
        Cursor cursor = db.rawQuery("SELECT count(*) FROM ProfilePic", null);

        if(cursor != null)
        {
            cursor.moveToFirst();
            if(cursor.getInt(0)== 0)
            {
                empty = true; //rows not null;
            }
            else
            {
                empty = false; // rows null;
            }
        }
        return empty;
    }

在sqlite数据库上返回太多数据警告

1 个答案: