我正在使用此代码更新 Firebase 上的数据，但它生成了新数据。我尝试了很多代码，除了最后一个字符外，它使用相同的键创建新的代码。

我将其用作推荐的 Firebase 站点，但它不起作用。我创建了一个新的，下次更新时:

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

这段代码也是一样

     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 push.getkey 并将其作为 ID 的问题 然后使用 push.setValue(masjed) 将使用相同的键 事实证明，当我使用它时， key 并不总是发生变化，它不存在，因此它创建了一个新的，帮助我的答案是切斯特答案

最佳答案

我希望这会有所帮助，在我的宠物项目中使用它(它不是清理解决方案，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

关于android - Firebase 更新正在制作除最后一个字符外具有相同 ID 的新版本，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/42559444/