使用选项从firebase数据库中检索随机数据

2 回答

• 0

  我一直在想如何做到这一点，我来到这里

  首先，您需要从 Database 获取所有数据，为此，请执行以下操作

  DatabaseReference ref= FirebaseDatabase.getInstance().getReference();
  
  
           //Since you dont have a parent for all the questions, you should reference to your url to get all the items
             ref.child("https://testapp-465fe.firebaseio.com/").addListenerForSingleValueEvent(new ValueEventListener() {
              @Override
              public void onDataChange(DataSnapshot dataSnapshot) {
  
                  questionList = new ArrayList<String>();                
                  // Result will be holded Here
                  for (DataSnapshot ql : dataSnapshot.getChildren()) {
                        questionList.add(String.valueOf(ql.geValue())); //We add all the results to an arraylist
  
                      }
          /* Now all the data from your database will be stored in the ArrayList, just go and get from there what you need, the answers, the questions, etc */
  

  我将留下关于如何从ArrayList获取内容的片段

  /*This method retrieves values from ArrayList using Iterator
       */
      public static void retrieveValuesFromList(List list)
      {
          Iterator itr = list.iterator();
          while(itr.hasNext())
          {
              System.out.println(itr.next()); //you can use a Log.e("Data: ",""+itr.next()); here
          }
      }
  

  其中 list 将是你的 ArrayList questionList

  retrieveValuesFromList(yourlist);
  

  希望这可以帮助

  快乐的编码！

  回复于 2024-05-12T01:47:21+08:00
• 0

  作为一个初学者，产生随机数可能是一个小混乱 . 所以一个简单的方法是在 millisecond 中获取 current time ，使其看起来像 11023049 (hour-min-sec-ms)

  由于每次生成时它都是唯一的，因此您需要一个函数来确定基于此唯一值检索的问题顺序 .

  要特定于您的问题，您可以在 millisecond 中获取时间，因为它是变化最快的变量并检索其数量等于毫秒的问题 . 或者有一些变化， odd millisecond values would fetch questions in reverse order, and even values would fetch in normal order .

  UPDATE 1 ：创建订单的一种简单方法是（这只会使起始问题随机化，下一个问题将按顺序排列，因此强烈建议使用其他方法）：

  public void ArrayList<Integer> getOrderTrivial(int lastQsNum)
  {
      System.out.println("Started");
      int currentSec = Integer.parseInt(new SimpleDateFormat("ss").format(new Date()));
      int count = 0;
      ArrayList<Integer> order = new ArrayList<Integer>();
      if (currentSec%2 == 0)
      {
          /*If the second value is even, the order of questions will be straight*/
          for (int i=currentSec; i<=lastQsNum; i++)
          {
              order.add(i);               
          }
          for (int i=0; i<=currentSec; i++)
          {
              order.add(i);               
          }
      }
      else
      {
          /*If the second value is even, the order of questions will be reverse*/
          for (int i=currentSec; i>=0; i--)
          {
              order.add(i);               
          }
          for (int i=lastQsNum; i>=currentSec; i--)
          {
              order.add(i);               
          }           
      }
      System.out.println("Complete");
      return order;
  }
  

  UPDATE 2: 更好的方法（推荐）

  public ArrayList<Integer> getOrder(int lastQsNum)
  {
      ArrayList<Integer> order = new ArrayList<>();
      int sec = Integer.parseInt(new SimpleDateFormat("ss").format(new Date()));
      int start = sec, down = sec - 1, up = sec + 1;      
      order.add(start);
      while (up<=lastQsNum && down>=0)
      {                       
          if (!order.contains(up)) order.add(up);         
          if (!order.contains(down)) order.add(down);
          up = up+1;
          down = down-1;                              
      }       
      if (up>lastQsNum)
      {
          System.out.println("Upper Limit Reaached: "+up+"-"+down);
          int mid = down/2;
          if (!order.contains(mid)) order.add(mid);
          up = mid+1;
          down = mid -1;          
      }
      else if (down<0)
      {
          System.out.println("Lower Limit Reaached: "+up+"-"+down);
          int mid = up + ((lastQsNum - up)/2);
          if (!order.contains(mid)) order.add(mid);
          up = mid+1;
          down = mid -1;
      }
      while (up<=lastQsNum && down>=0)
      {
          if (sec%3 == 0)
          {
              if (!order.contains(up)) order.add(3, up);          
              if (!order.contains(down)) order.add(6, down);
          }
          else if (sec%5 == 0)
          {
              if (!order.contains(up)) order.add(5, up);          
              if (!order.contains(down)) order.add(10, down);
          }
          else if (sec%7 == 0)
          {
              if (!order.contains(up)) order.add(7, up);          
              if (!order.contains(down)) order.add(14, down);
          }
          else if (sec%11 == 0)
          {
              if (!order.contains(up)) order.add(11, up);         
              if (!order.contains(down)) order.add(22, down);
          }
          else if (sec%2 == 0)
          {
              if (!order.contains(up)) order.add(0, up);          
              if (!order.contains(down)) order.add(0, down);  
          }
          else
          {
              if (!order.contains(up)) order.add(up);         
              if (!order.contains(down)) order.add(down);
          }           
          up = up+1;
          down = down-1;          
      }
      return order;
  }
  

  两个方法的返回对象是ArrayList . 在Activity的 onCreate() 存根中，您应该运行此函数并创建一个ArrayList . 就像是：

  //Global Variables
  ArrayList<Integer> order;
  int maxQuestions = 200;
  
  @Override
  protected void onCreate(Bundle savedInstanceState)
  {
      super.onCreate(savedInstanceState);
      setContentView(R.layout.activity_main);
      order = getOrderTrivial(maxQuestions);
  }
  

  一旦创建了 ArrayList order ，您就可以串行遍历 ArrayList 并获取问题编号并使用它来从 Firebase Database （来自ValueEventListener）获取问题 .

  回复于 2024-05-12T01:47:21+08:00