添加for循环时，Firebase addListenerForSingleValueEvent将不起作用

Question

我想登录两种不同的用户类型。所以我这样编码。但是addListenerForSingleValueEvent中的for循环在此不起作用，请帮助。它好像不会进入for循环。请帮助我在这个登录卡。JAVA

    login.setOnClickListener(new View.OnClickListener() {
        @Override
        public void onClick(View v) {
            performLogin();
        }
    });
}
private String email,password;
private void performLogin() {
    email=emaillog.getText().toString().trim();
    password=passwordlog.getText().toString().trim();
    if(!Patterns.EMAIL_ADDRESS.matcher(email).matches()) {
        Toast.makeText(this, "Invalid Email Address", Toast.LENGTH_SHORT).show();
        return;
    }
    if(TextUtils.isEmpty(password)) {
        Toast.makeText(this, "Password should not be empty", Toast.LENGTH_SHORT).show();
        return;
    }
    progressDialog.setMessage("Logging In....");
    progressDialog.show();
    firebaseAuth.signInWithEmailAndPassword(email,password)
            .addOnSuccessListener(new OnSuccessListener<AuthResult>() {
                @Override
                public void onSuccess(@NonNull AuthResult authResult) {
                    //logged in successfully
                    checkOnline();
                }
            })
            .addOnFailureListener(new OnFailureListener() {
                @Override
                public void onFailure(@NonNull Exception e) {
                    //logging in failed
                    progressDialog.dismiss();
                    Toast.makeText(Login.this, ""+e.getMessage(), Toast.LENGTH_SHORT).show();

                }
            });
}

private void checkOnline() {
    //after loggin in
    progressDialog.setMessage("Checking User....");
    HashMap<String, Object> hashMap = new HashMap<>();
    hashMap.put("online","true");
    //update value to db
    DatabaseReference ref = FirebaseDatabase.getInstance().getReference("Users");
    ref.child(firebaseAuth.getUid()).updateChildren(hashMap)
            .addOnSuccessListener(new OnSuccessListener<Void>() {
                @Override
                public void onSuccess(@NonNull Void unused) {
                    //Update by checking user
                    checkUserType();
                }
            })
            .addOnFailureListener(new OnFailureListener() {
                @Override
                public void onFailure(@NonNull Exception e) {
                    //update unsuccessfully
                    progressDialog.dismiss();
                    Toast.makeText(Login.this, ""+e.getMessage(), Toast.LENGTH_SHORT).show();
                }
            });
}

private void checkUserType() {
    progressDialog.setMessage("Checking UserType....");
    //if the user is seller,then go into seller dashboard
    // if the user is customer, then go into customer dashboard
    DatabaseReference ref = FirebaseDatabase.getInstance().getReference("Users");
    progressDialog.setMessage("Checking UserType1....");
    ref.orderByChild("uid").equalTo(firebaseAuth.getUid())
            .addListenerForSingleValueEvent(new ValueEventListener() {
                @Override
                public void onDataChange(@NonNull DataSnapshot snapshot) {
                    **for (DataSnapshot ds: snapshot.getChildren()){
                        String accounttype =""+ds.child("accounttype").getValue();
                        if(accounttype == "Seller"){
                            startActivity(new Intent(Login.this,SellerDahboard.class));
                        }
                        else{
                            startActivity(new Intent(Login.this,SellerDahboard.class));
                        }
                    }**
                }

                @Override
                public void onCancelled(@NonNull DatabaseError error) {
                }
            });

火基

Answer 1

第一步是忽略潜在的错误，并在取消后实现。

至少应该是这样的：

public void onCancelled(@NonNull DatabaseError databaseError) { 
    throw databaseError.toException(); 
}

或：

public void onCancelled(@NonNull DatabaseError databaseError) { 
    Log.e("Firebase", databaseError.toException());
}

如果这显示了来自数据库的异常，我建议搜索该错误消息，因为它可能已经被覆盖过了。