Firebase实时数据库记录只有一个孩子

Question

对于特定用途，我只想在实时数据库中保存一个元素，并且在修改时更新它，我不需要将所有更改保存为子项。 我有这段代码，但它创建了孩子。

<script type="module">    
import {initializeApp} from "https://www.gstatic.com/firebasejs/9.6.6/firebase-app.js";
import {
    getDatabase,
    set,
    ref,
    push,
    child,
    onValue,
    onChildAdded
} from "https://www.gstatic.com/firebasejs/9.6.6/firebase-database.js";    

const firebaseConfig = {
  apiKey: "...",
  authDomain: "....firebaseapp.com",
  databaseURL: "....firebaseio.com",
  projectId: "...",
  storageBucket: "...",
  messagingSenderId: "...",
  appId: "..."
 };

// Initialize Firebase
const app = initializeApp(firebaseConfig);
const database = getDatabase(app);    

submit.addEventListener('click', (e) => {
    var message = document.getElementById('message').value;
    //var name = myName;

    const id = push(child(ref(database), 'messages')).key;

    set(ref(database, 'messages/' + id), {
        //name: name,
        message: message
    });
    document.getElementById('message').value = "";
    //alert('message has sent');

});

const dbRef = ref(getDatabase());
const newMsg = ref(database, 'messages/');
onChildAdded(newMsg, (data) => {        
        var divData = data.val().message;
        let h1 = document.querySelector("h1");
        h1.textContent = divData;
});

谢谢。

Answer 1

如果你不想创建一个子节点，你可以简单地写messages ：

set(ref(database, 'messages'), {
    message: message
});