Firebase 实时数据库 setValue() 不起作用

如何解决Firebase 实时数据库 setValue() 不起作用

我正在尝试将有关用户的一些基本信息写入 Firebase 实时数据库，但在运行 setValue() 时，控制台中没有任何反应，经过数小时的尝试后，我仍然无法找出问题所在。

>

这是调用 registerUser 函数的片段中的代码：

registerButton.setonClickListener {

        firstNameLayout.error = null
        lastNameLayout.error = null
        emailFieldLayout.error = null
        passwordFieldLayout.error = null

        var fieldCheck = true

        if(TextUtils.isEmpty(firstNameField.text)) {
            firstNameLayout.error = "Please Enter First Name";
            fieldCheck = false
        }
        if (TextUtils.isEmpty(lastNameField.text)) {
            lastNameLayout.error = "Please Enter Last Name"
            fieldCheck = false
        }
        if(TextUtils.isEmpty(emailField.text)) {
            emailFieldLayout.error = "Please Enter Email Address";
            fieldCheck = false
            // Todo: Make sure format is correct
        }
        if(TextUtils.isEmpty(passwordField.text)){
            passwordFieldLayout.error = "Please Choose a Password"
            fieldCheck = false
            // Todo: Stricter password requirements 
        }
        if(!fieldCheck) {
            return@setonClickListener
        }

        registerButton.startAnimation()

        val fName = firstNameField.text.toString().trim()
        val lName = lastNameField.text.toString().trim()
        val email = emailField.text.toString().trim()
        val password = passwordField.text.toString().trim()

        try {
            registerButton.doneLoadingAnimation(2,bitmap)
            (activity as LoginRegister).registerUser(email,password,fName,lName)

        } catch (e: Exception) {
            Toast.makeText(activity,e.message,Toast.LENGTH_SHORT).show()
            registerButton.revertAnimation()
        }

    }

这是来自片段父活动的 registerUser 函数体：

fun registerUser(email: String,password: String,fName: String,lName: String) {
    //Registrerer med Firebase Authentication
    auth.createuserWithEmailAndPassword(email,password)
        .addOnCompleteListener(this) { task ->
            if (task.isSuccessful) {
                // Sign in success,update UI with the signed-in user's information
                Log.d(TAG,"createuserWithEmail:success")
                val user : FirebaseUser = task.result!!.user!!
                val uid: String = user.uid
                val dbUser = User(fName,lName,email)

                writeNewUser(uid,dbUser)

                val intent = Intent(this@LoginRegister,MainActivity::class.java)
                intent.flags = Intent.FLAG_ACTIVITY_NEW_TASK or Intent.FLAG_ACTIVITY_CLEAR_TASK
                intent.putExtra("user_email",email)
                intent.putExtra("user_first_name",fName)
                intent.putExtra("user_last_name",lName)
                startActivity(intent)
                finish()

            } else {
                // If sign in fails,display a message to the user.
                Log.w(TAG,"createuserWithEmail:failure",task.exception)
                Toast.makeText(baseContext,"Authentication Failed.",Toast.LENGTH_SHORT).show()
            }
        }



}

这是 writeNewUser 函数体

fun writeNewUser(userId: String,user: User) {
    database = FirebaseDatabase.getInstance().getReference("Users").child(userId)
    database.child("users").child(userId).setValue(user)
}

User 对象是从这个 kotlin 类实例化的：

data class User(val fName: String? = null,val lName: String? = null,val email: String? = null) {
val firstName = fName?.capitalize(Locale.ROOT)
val lastName = lName?.capitalize(Locale.ROOT)
val emailAddress = email

}

Firebase：

Firebase nodes

有人知道这背后的原因吗？

解决方法

fun writeNewUser(userId: String,user: User) {
    database = FirebaseDatabase.getInstance().getReference("Users").child(userId)
    database.child("users").child(userId).setValue(user)
}

我猜你在上面的代码片段中复制了几个子节点，因为这意味着一组用户有一个用户，而这个特定用户有一组用户，没有意义，所以你可以删除这些重复项：

fun writeNewUser(userId: String,user: User) {
    database = FirebaseDatabase.getInstance().getReference("Users").child(userId)
    database.setValue(user)
}

更新：

firebase 节点区分大小写，因此将“Users”更改为小写的“users”

fun writeNewUser(userId: String,user: User) {
    database = FirebaseDatabase.getInstance().getReference("users").child(userId)
    database.setValue(user)
}

,

我解决了这个问题，我使用的不是默认的数据库位置，我没有发现我需要在 getInstance() 方法中传递数据库 URL 的事实。传递正确的数据库 url 修复了问题

,

将数据库的 URL 放在 getInstance() 中是我的解决方案，现在它完美无缺

示例：

private val database = FirebaseDatabase.getInstance("https://conect-c91b3-default-rtdb.firebaseio.com/").getReference()

// to test if is working
database.child("users").setValue("test")