尝试从SQLite

Question

单击按钮时，我想从SQLite获取数据ID。不幸的是我的应用程序崩溃了。当我尝试从SQLite中检索数据ID时会发生这种情况。

logcat的

05-17 01:00:54.873  16093-16093/? E/AndroidRuntime﹕ FATAL EXCEPTION: main
android.database.CursorIndexOutOfBoundsException: Index -1 requested, with a size of 1
at android.database.AbstractCursor.checkPosition(AbstractCursor.java:424)
at android.database.AbstractWindowedCursor.checkPosition(AbstractWindowedCursor.java:136)
at android.database.AbstractWindowedCursor.getInt(AbstractWindowedCursor.java:68)
at com.example.tony.easyactivity.MainActivity$1.onClick(MainActivity.java:49)

第49行

int id = data.getInt(data.getColumnIndex("ID"));

非常感谢任何帮助。

我尝试过：

btnAddData.setOnClickListener(new View.OnClickListener() {
                @Override
                public void onClick(View v) {
                    boolean isInserted = myDb.insertData(editName.getText().toString());
                    if (isInserted) {
                        Cursor data = myDb.getID(editName.getText().toString());
                        int id = data.getInt(data.getColumnIndex("ID"));
                        Toast.makeText(getApplicationContext(), id + "", Toast.LENGTH_LONG).show();
                    }
                    else
                    {
                        myDb.addData(editName.getText().toString());
                        Toast.makeText(getApplicationContext(), "Not yet inserted ", Toast.LENGTH_LONG).show();
                    }
                }
            });
        }

DatabaseHelper

public Cursor getID(String info) {
            SQLiteDatabase db = this.getWritableDatabase();
            Cursor c = db.rawQuery("Select ID from " + TABLE_NAME1 + " WHERE " + COL3 + "=?", new String[]{info});
            return c;
        }

        public boolean insertData(String name) {
            SQLiteDatabase db = this.getReadableDatabase();
            Cursor c = db.rawQuery("Select * from " + TABLE_NAME1 + " WHERE " + COL3 + "=?", new String[]{name});
            if (c.getCount() > 0) {
                return true;
            } else
                return false;
        }

        public void addData(String name) {
            SQLiteDatabase db = this.getWritableDatabase();
            ContentValues cv=new ContentValues();
            cv.put(COL3, name);
            db.insert(TABLE_NAME1, null, cv);
        }

Answer 1

您需要将Cursor移动到第一个条目，然后才能使用moveToFirst()方法访问其数据。所以在onClickListener()：

Cursor data = myDb.getID(editName.getText().toString());
data.moveToFirst();
int id = data.getInt(data.getColumnIndex("ID"));