为什么在es6中需要构造函数？

Question

为什么当我在test id中编写'new ServerNotificationApi'时不会调用构造函数,因为我new ServerNotificationApi.constructor()工作,但我无法理解为什么当我写 new ServerNotificationApi单元测试时出现错误'TypeError:_serverNotifications.default不是构造函数'

类

class ServerNotificationApi {
        constructor() {
            SignalR.initConnection(url.serverNotificationHubName)
        }

        subscribe = callback => SignalR.subscribe(url.entityChanged, url.serverNotificationHubName, callback);

        unsubscribe = callback => SignalR.unsubscribe(url.entityChanged, url.serverNotificationHubName, callback);
    }

    export default new ServerNotificationApi()

测试

 it('constructor should call signalR method \'initConnection\'', () => {
        sinon.stub(SignalR, 'initConnection')

        new ServerNotificationApi.constructor()

    SignalR.initConnection.calledWith(url.serverNotificationHubName).should.be.true

        SignalR.initConnection.restore()
    })

Answer 1

export default new ServerNotificationApi()
               ???

您正在导出类的实例,而不是类本身.你基本上是这样做的:

let foo = new ServerNotificationApi();
new foo();

哪,是的,不起作用.摆脱new中export.