android – Firebase更新正在创建新的与除了最后一个字符相同的id

我正在使用这个代码来更新Firebase上的数据,但它是新的.我尝试了很多代码,并且使用相同键的新的代码,除了最后一个字符.

我用这个作为Firebase网站推荐,但它不工作.我创建了一个新的下一次更新：

Map<String,Object> childUpdate = new HashMap<>();
childUpdate.put("/masjeds/" + masjed.getId(),masjed.toMap());
reference.updateChildren(childUpdate);

而这段代码也是一样的

final FirebaseDatabase database = FirebaseDatabase.getInstance();
    masjeds = database.getReference("masjeds");

        reference.child(masjed.getId()).setValue(masjed,new DatabaseReference.CompletionListener() {
        @Override
        public void onComplete(DatabaseError databaseError,DatabaseReference databaseReference) {
         //   Toast.makeText(MyMasjedsActivity.this,databaseError.getMessage(),Toast.LENGTH_SHORT).show();
        }
    });

Masjed类是纯java对象

public class Masjed {
private String userID;
private String id;
private String name;
private String address;
private String phone;
private boolean matloopEmam;

public String getId() {
    return id;
}

public void setId(String id) {
    this.id = id;
}

public String getUserID() {
    return userID;
}

public void setUserID(String userID) {
    this.userID = userID;
}

public boolean isMatloopEmam() {
    return matloopEmam;
}

public void setMatloopEmam(boolean matloopEmam) {
    this.matloopEmam = matloopEmam;
}


public Masjed(String name,String address,String phone) {
    this.name = name;
    this.address = address;
    this.phone = phone;
}

public Masjed() {

}

public String getName() {
    return name;
}

public void setName(String name) {
    this.name = name;
}

public String getAddress() {
    return address;
}

public void setAddress(String address) {
    this.address = address;
}

public String getPhone() {
    return phone;
}

public void setPhone(String phone) {
    this.phone = phone;
}

public Map<String,Object> toMap() {
    Map<String,Object> map = new HashMap<>();
    map.put("name",name);
    map.put("address",address);
    map.put("phone",phone);
    map.put("id",id);
    map.put("userID",userID);
    return map;
}

}

id是我假设push.getkey并将其作为ID的问题
那么使用push.setValue(masjed)将使用相同的键
事实证明,当我使用它的dosnt存在时,它并不总是Case的改变,所以它创建了一个帮助我的答案是Chester Answer

解决方法

我希望这有助于,在我的宠物项目中使用它(这不是清洁的解决方案,但它的工作原理)：

masjeds = database.getReference("masjeds");

ValueEventListener listener = new ValueEventListener() {
    @Override
    public void onDataChange(DataSnapshot dataSnapshot) {
        for (DataSnapshot dataSnapshot1 : dataSnapshot.getChildren()) { 
                masjedKey = dataSnapshot1.getKey(); // This is a member variable
                masjed = dataSnapshot1.getValue(Masjed.class); // This is a member variable
        }
    }

    @Override
    public void onCancelled(DatabaseError databaseError) {

    }
};
masjeds.orderByChild("id").equalTo(masjed.getId()).addValueEventListener(listener);

//Use your Map to update each value this is just an example
masjeds.child(masjedKey).child("name").setValue("John"); //you can add event listener if you want to see if it's completed,but it works without the events