Question

如何将websocket属性onopen回调委托给外部对象的方法，外部对象为this，而不是websocket为this ？（beaceuse websocket没有函数showAlert

换句话说：如何在webscoket上呼叫我的MyClient.showConnected？

function MyClient() {
    this.websocket = {};
}

MyClient.prototype.showAlert = function( ) {
    alert("Connected!");
};

MyClient.prototype.showConnected = function( evt ) {
    this.showAlert();
};

MyClient.prototype.connect = function( url ) {
    this.websocket = new WebSocket(url);
    this.websocket.onopen =  this.showConnected;
}

这种方式产生this.showAlert is not a function。用法：

var client = new MyClient();
client.connect("ws://.......");

这也是不工作：

this.websocket.onopen = function(evt) { this.onOpen(evt); };

Answer 1

function MyClient() {
    this.websocket = {};
}

MyClient.prototype.showAlert = function( ) {
    alert("Connected!");
};

MyClient.prototype.connect = function( url ) {
    this.websocket = new WebSocket(url);
    this.websocket.onopen = this.showAlert.bind(this);
}

Answer 2

这是你想要的吗？

function MyClient() {
    this.websocket = {};
}

MyClient.prototype.showAlert = function( ) {
    alert("Connected!");
};

MyClient.prototype.connect = function( url ) {
    this.websocket = new WebSocket(url);
    var that = this;
    // that is hidden in connect namespace, some people use _this instead
    this.websocket.onopen = function( evt ) {
        that.showAlert();
    };
}

我认为您的问题是onopen是事件。如果是事件，JS总是更改this。然后，最好的选择是使用某个命名空间中的变量并在其中保存this。

如何将JavaScript回调设置为外（包装）对象的方法？

2 个答案: