Firebase更新是新的一个除了最后一个字符之外具有相同的ID

Question

我正在使用此代码更新Firebase上的数据,但它会生成新的代码.我尝试了很多代码,除了最后一个字符之外,它使用相同的密钥制作新代码.

我使用此作为Firebase网站推荐,但它不起作用.我创建了一个新的,并在下次更新时:

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

这段代码完成了同样的事情

     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 Class是普通的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是我假设的那个问题我假设push.getkey并把它作为ID然后使用push.setValue(masjed)将使用相同的密钥它结果是它并不总是Case当我使用它时密钥更改dosnt存在所以它创造了新的答案,帮助我切斯特答案

Answer 1

我希望这有帮助，在我的宠物项目中使用它（它不是清洁解决方案，但它有效）：

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