firebase检索最高100分 [英] firebase retrieve highest 100 score

问题描述

这是我的firebase的屏幕截图

i试图检索Firebase数据库中最高的100分
i使用此代码向firebase添加新节点

  Map< String，String> post1 = new HashMap< String，String>（）; 
 post1.put（name，name）; 
 post1.put（score，score）; 
 myRef.push（）。setValue（post1）; 
  

这是我用来检索最高100分的代码，代码的作品，但它不是检索最高的100分）

 查询queryRef = myRef.orderByChild（score）。limitToFirst 100）; （DataSnapshot postSnapshot：dataSnapshot.getChildren（））返回一个新的DataEventListener对象，并将其添加到DataSnapshot的DataSnapshot对象中。 {
 Score score = postSnapshot.getValue（Score.class）; 
 Log.d（test，values is+ score.getName（）++ score.getScore（））; 
 
 
 $ b @Override 
 public void onCancelled（DatabaseError databaseError）{
 
 
}）; 
  

解决方案

Firebase查询始终以升序排列。因此，您需要获取 last 100，而不是前100个。

 查询queryRef = myRef.orderByChild（score）。limitToLast（100）; 
  

然后在客户端，您需要将这些项目反向。

另外，你可以添加一个倒置的属性到你的项目 invertedScore：-99 。如果你这样做的话，你可以按倒数得分来排序，而不必颠倒数组。

这个场景已经被经常覆盖了。我强烈建议你研究一下这些：

• 数据在火力点排序 （就在昨天）


• 以降序显示帖子
-def-sorting-in-android / 34158197＃34158197> Firebase数据存储在Android中排序

>

this is a screen shot of my firebase i am trying to retrieve the highest 100 score in firebase database i am using this code to add new node to firebase

 Map<String, String> post1 = new HashMap<String, String>();
        post1.put("name",name);
        post1.put("score",score);
        myRef.push().setValue(post1);

and this is the code i am using to retrieve the highest 100 score which doesn't work (the code works but it is not retrieving the highest 100 score)

 Query queryRef = myRef.orderByChild("score").limitToFirst(100);
        queryRef.addValueEventListener(new ValueEventListener() {
            @Override
            public void onDataChange(DataSnapshot dataSnapshot) {

                for (DataSnapshot postSnapshot: dataSnapshot.getChildren()) {
                    Score score=postSnapshot.getValue(Score.class);
                    Log.d("test"," values is " + score.getName()  + " " + score.getScore());
                }
            }

            @Override
            public void onCancelled(DatabaseError databaseError) {

            }
        }); 

解决方案

Firebase queries are always in ascending order. So you'll need to get the last 100, instead of the first 100.

Query queryRef = myRef.orderByChild("score").limitToLast(100);

Then client-side you'll need to reverse the items.

Alternatively you can add a inverted property to your items invertedScore: -99. If you do that, you can order by that inverted score and won't have to reverse the array.

This scenario has been covered frequently before. I highly recommend you study some of these:

• data sorting in firebase
• Firebase Leaderboard, best Implmentation. SQL Joins and orderby (as recently as yesterday)
• Display posts in descending posted order
• Firebase Data Desc Sorting in Android
• Swift - How to create Sort query as Descending on Firebase?

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