Question

所以这些是我的表格;

我以这样的方式保存我的数据：Workout_ID（第二个图）引用了上图中的Workout_ID。

我需要什么 - 在显示我的桌子时，我如何'查找'Workout_ID 2是什么并输出字符串？

在SQLite中完成，这可能吗？

编码表列名称：

date_id = date_of_workout_id
date = date_of_workout
workout_name = workout_name
date_of_workout = DateofWorkout
workout_table = WorkoutTable
workout_id(date's one) = name_of_workout
workout_id(WorkoutTables's one) = workout_id

public String test（String WorkoutSelectedNameInfo）{

    // TODO Auto-generated method stub
    String Weight = "";
    open();
    ourDatabase = ourhelper.getReadableDatabase();



    Cursor c = ourDatabase.rawQuery("SELECT date_of_workout_id, 
   date_of_workout, workout_name FROM DateofWorkout JOIN WorkoutTable ON 
 DateofWorkout.name_of_workout = WorkoutTable.workout_id", null);
    int iWeight = c.getColumnIndex(KEY_WORKOUT_NAME);
             while(c.moveToNext())
        {
            Weight = Weight + c.getString(iWeight) + "\n";
        }


    c.close();
    ourDatabase.close();

    System.out.println(Weight);

    return Weight;



}

Answer 1

更新的答案：

尝试更改

Cursor c = ourDatabase.rawQuery（“SELECT date_of_workout_id，
date_of_workout，workout_name FROM DateofWorkout JOIN WorkoutTable ON DateofWorkout.name_of_workout = WorkoutTable.workout_id“，null）;

到

Cursor c = ourDatabase.rawQuery（“SELECT date_of_workout_id，
date_of_workout，workout_name FROM DateofWorkout JOIN WorkoutTable ON DateofWorkout.workout_id = WorkoutTable.workout_id “，null）;

问题可能是因为您正在比较 workout_id 中的 name_of_workout 。

也许是一个简单的内部联接？

select dow.date_id, dow.date, w.name
from workout w, date_of_workout dow
where w.workout_id = dow.workout_id

Answer 2

一个非常简单的JOIN可以做到这一点;

SELECT date_id, date, workout_name
FROM date_of_workout 
JOIN workout_table 
  ON date_of_workout.workout_id = workout_table.workout_id

Answer 3

SELECT date_ID, date, workout_name
FROM workouts, dates
WHERE workouts.workout_ID = dates.workout_ID

在SQLite上执行查找

3 个答案: