Warning: file_get_contents(/data/phpspider/zhask/data//catemap/3/android/189.json): failed to open stream: No such file or directory in /data/phpspider/zhask/libs/function.php on line 167

Warning: Invalid argument supplied for foreach() in /data/phpspider/zhask/libs/tag.function.php on line 1116

Notice: Undefined index: in /data/phpspider/zhask/libs/function.php on line 180

Warning: array_chunk() expects parameter 1 to be array, null given in /data/phpspider/zhask/libs/function.php on line 181
Java 在Firebase中使用orderByChild时,如何获取节点名?_Java_Android_Firebase_Firebase Realtime Database_Nosql - Fatal编程技术网

Java 在Firebase中使用orderByChild时,如何获取节点名?

Java 在Firebase中使用orderByChild时,如何获取节点名?,java,android,firebase,firebase-realtime-database,nosql,Java,Android,Firebase,Firebase Realtime Database,Nosql,这是我的密码: @Override public View onCreateView(LayoutInflater inflater, ViewGroup container, Bundle savedInstanceState) { Query queryRef = mdatabase.child("Users").orderByChild("age"); queryRef.addValueEventListener(new

这是我的密码:

@Override
public View onCreateView(LayoutInflater inflater, ViewGroup container,
                         Bundle savedInstanceState) {

    Query queryRef  = mdatabase.child("Users").orderByChild("age");
    queryRef.addValueEventListener(new ValueEventListener() {
        @Override
        public void onDataChange(DataSnapshot snapshot) {
            for (DataSnapshot chatSnapshot: snapshot.getChildren()) {
                User users = chatSnapshot.getValue(User.class);
                String name = users.getName();
                System.out.println(name);
            }
        }

        @Override
        public void onCancelled(DatabaseError databaseError) {

        }
    });        // Inflate the layout for this fragment
    return inflater.inflate(R.layout.fragment_home, container, false);
}
正如你们可能猜到的,这段代码是用来显示用户名的。但我实际上想做的是,基于相同的查询结果,我想显示他们的用户ID,而不是名称。我尝试使用这样的类来获取用户id:

public class User{
private String gender;
private Long age;
private String email;
private Long height;
private Long weight;
private String name;
private String type;
private String userId;

public User()
{

}
public User(String gender,Long age,String email,Long height,Long weight,String name,String type,String userId)
{
    this.gender=gender;
    this.age=age;
    this.email=email;
    this.height=height;
    this.weight=weight;
    this.name=name;
    this.type=type;
    this.userId = FirebaseAuth.getInstance().getCurrentUser().getUid();
}

public String getId()
{
    return userId;
}

public String getGender()
{
    return gender;
}

public Long getAge()
{
    return age;
}

public String getEmail()
{
    return email;
}

public Long getHeight()
{
    return height;
}

public Long getWeight()
{
    return weight;
}

public String getName()
{
    return name;
}

public String getType()
{
    return type;
}}
但这会返回当前用户id 3次,因为我的数据库中有3个用户,而不是单个用户id。 这是我的数据库:

每个用户都有这么多值

我尝试的另一种方法是使用getKey(),如下所示:

@Override
public View onCreateView(LayoutInflater inflater, ViewGroup container,
                     Bundle savedInstanceState) {

Query queryRef  = mdatabase.child("Users").orderByChild("age");
queryRef.addValueEventListener(new ValueEventListener() {
    @Override
    public void onDataChange(DataSnapshot snapshot) {
        for (DataSnapshot chatSnapshot: snapshot.getChildren()) {
            String name = snapshot.getKey();
            System.out.println(name);
        }
    }

    @Override
    public void onCancelled(DatabaseError databaseError) {

    }
});        // Inflate the layout for this fragment
return inflater.inflate(R.layout.fragment_home, container, false);
}
但这会返回节点“Users”3次,而不是用户id

有人能帮我吗?我想显示他们的用户ID。即:

OMBACIWG3FOURXY8SEQD8TAQL12,kZllc0GL1Vfbitt73WHWwJuXvXC2,lxVykDo9GQRcFmb0u1QZ0WCc9lg2 (根据图片)


非常感谢您的帮助。谢谢。

您在错误的
DataSnapshot
处调用了
getKey()
方法。 这就是你想要的

@Override
public void onDataChange(DataSnapshot snapshot) {
    for (DataSnapshot chatSnapshot: snapshot.getChildren()) {
        String key = chatSnapshot.getKey();
        System.out.println(key);
    }
}
编辑:将userId设置为
User

for (DataSnapshot chatSnapshot: snapshot.getChildren()) {
    User user = chatSnapshot.getValue(User.class);
    user.setUserId(chatSnapshot.getKey());
}

非常感谢。现在它工作得很好!:D只是出于好奇,有没有办法使用我在问题中提到的类来获取用户ID?这是不可能的,你必须手动设置它。我已经更新了我的答案@雅各布斯汀