Question

我的return Car;存在问题我认为它会返回“KEY_IMAGE”字符串。

KEY_IMAGE是一个字符串，将放在BitmapFactory中：

byte[] logoImage = getLogoImage(IMAGEURL);

private byte[] getLogoImage(String url){
     try {
             URL imageUrl = new URL(url);
             URLConnection ucon = imageUrl.openConnection();

             InputStream is = ucon.getInputStream();
             BufferedInputStream bis = new BufferedInputStream(is);

             ByteArrayBuffer baf = new ByteArrayBuffer(500);
             int current = 0;
             while ((current = bis.read()) != -1) {
                     baf.append((byte) current);
             }

             return baf.toByteArray();
     } catch (Exception e) {
             Log.d("ImageManager", "Error: " + e.toString());
     }
     return null;
}

logoImage.setImageBitmap(BitmapFactory.decodeByteArray( currentAccount.accImage, 
        0,currentAccount.accImage.length));

我的方法有错误：

public Car getCurrentCar() {
    SQLiteDatabase db       =   getWritableDatabase();
    String sql              =   "SELECT " + KEY_ID + "," + KEY_IMAGE + " FROM " + TABLE_CARS + "ORDER BY RANDOM() LIMIT 1";
    Cursor cursor           =   db.rawQuery(sql, new String[] {});

    if(cursor.moveToFirst()){
    Car car_current = new Car(Integer.parseInt(cursor.getString(0)), cursor.getString(1), cursor.getString(2),cursor.getString(3));
    }
    if (cursor != null && !cursor.isClosed()) {
        cursor.close();
    }
    db.close();
    if(cursor.getCount() == 0){
        return null;
    } else {
        return Car;
    }
}

回车;给了我错误：

“表达预期”和“找不到符号变量Car”。

有人能够解释为什么我不能回车吗？

编辑：

public Car getCurrentCar() {

    SQLiteDatabase db       =   getWritableDatabase();
    String sql              =   "SELECT " + KEY_ID + "," + KEY_IMAGE + " FROM " + TABLE_CARS + "ORDER BY RANDOM() LIMIT 1";
    Cursor cursor           =   db.rawQuery(sql, new String[] {});
    Car car = null;

    try {
        if (cursor.moveToFirst()) {
            car = new Car(Integer.parseInt(cursor.getString(0)), cursor.getString(1), cursor.getString(2), cursor.getString(3));
        }
    }
    finally {
        if (cursor != null && !cursor.isClosed()) {
            cursor.close();
        }
            db.close();
    }
    return car;
}

Answer 1

找不到符号：变量Car

这意味着编译器无法找到名为Car的变量。

return Car;

Car是一个类，而不是一个对象。您必须返回一个对象（或变量）。

我认为这就是你的意思

return car_current;

但你必须在if之外声明该变量：

Car car_current = null;
if(cursor.moveToFirst()){
    car_current = new Car(Integer.parseInt(cursor.getString(0)), cursor.getString(1), cursor.getString(2),cursor.getString(3));
}

Answer 2

您已在另一个car_current内宣布if，其范围不同。该变量在if块之外没有范围。如果你想在`if block。

之外使用它，你应该在if之外声明它

其次，您正在返回Car这是一个类，因此它无效。相反，您可能想要返回car_current我想。

Answer 3

这是Java 101。

试试这样：

// Why not pass in a Car id? 
public Car getCurrentCar() {
    // Bad form here.  Externalize these.
    SQLiteDatabase db       =   getWritableDatabase();
    String sql              =   "SELECT " + KEY_ID + "," + KEY_IMAGE + " FROM " + TABLE_CARS + "ORDER BY RANDOM() LIMIT 1";
    Cursor cursor           =   db.rawQuery(sql, new String[] {});

    Car car = null;
    if(cursor.moveToFirst()) {
        car = new Car(Integer.parseInt(cursor.getString(0)), cursor.getString(1), cursor.getString(2),cursor.getString(3));
    }
    // Wrong - close in a finally block.
    if (cursor != null && !cursor.isClosed()) {
        cursor.close();
    }
    db.close(); // Wrong - close in a finally block.
    return car;
}

Answer 4

我认为你需要返回car_current而不是Car。但是变量car_current不在范围内，因此您应该在if语句之外定义它。试试这个：

Car car_current = null;
if(cursor.moveToFirst()){
     car_current = new Car(Integer.parseInt(cursor.getString(0)), cursor.getString(1), cursor.getString(2),cursor.getString(3));

}
//after some code you could return only car_current without control
return car_current;

无法返回值 - 表达式预期？

4 个答案: