javascript中的继承问题。超级表达式必须为null或函数

Question

这是我的代码。不知道如何使用。

import firebase from "firebase";

export default class database extends firebase{

constructor（）{

super(
      config = {
        apiKey: "AIzaSyB0y847OMwZQ2BJqz8yQjjWB55YPPUk7lM",
        authDomain: "mytodo-faa53.firebaseapp.com",
        databaseURL: "https://mytodo-faa53.firebaseio.com",
        storageBucket: "mytodo-faa53.appspot.com",
        messagingSenderId: "86914291874"
      },
      fireApp = firebase.initializeApp(config),
      database = fireApp.database(),
      databaseRef = database.ref().child('Todos')
  )

 }

}

Answer 1

将firebase放在花括号中：

import {firebase} from "firebase";

稍后您可能会收到另一个错误，因为super函数应该在继承类的构造函数中使用：

export default class database extends firebase {
    constructor() {
       super (
          config = {
            apiKey: "AIzaSyB0y847OMwZQ2BJqz8yQjjWB55YPPUk7lM",
            authDomain: "mytodo-faa53.firebaseapp.com",
            databaseURL: "https://mytodo-faa53.firebaseio.com",
            storageBucket: "mytodo-faa53.appspot.com",
            messagingSenderId: "86914291874"
          },
          fireApp = firebase.initializeApp(config),
          database = fireApp.database(),
          databaseRef = database.ref().child('Todos')
       );
    }
}