将 android 个应用程序连接到 sockjs 服务器

Question

我正在开发原生 android 应用程序，我必须将我的应用程序连接到 Sockjs 服务器。我尝试连接 Socket.IO、Autobahn、WebSockets，但未连接。 android 是否有任何 sockjs 客户端？请帮忙，谢谢。

Socket.IO :

private void connectToSocket(String serverUrl, String accessKey) {
    IO.Options opts = new IO.Options();
    opts.forceNew = true;
    opts.reconnection = true;
    //opts.query = "accessKey="+accessKey;
    String host = serverUrl;

    try {
        mSocket = IO.socket(host, opts);
    } catch (URISyntaxException e) {
        e.printStackTrace();
    }

    mSocket.on(Socket.EVENT_CONNECT_ERROR, onConnectError);
    mSocket.on(Socket.EVENT_CONNECT_TIMEOUT, onConnectError);
    mSocket.on(Socket.EVENT_CONNECT, onConnect);
    mSocket.on("new message", onNewMessage);
    mSocket.connect().emit("accessKey", accessKey);
    if (mSocket.connected()) {
        Log.d("SocketMsg: ", "Connected");
    }

}

private Emitter.Listener onConnect = new Emitter.Listener() {

    @Override
    public void call(final Object... args) {
        MainActivity.this.runOnUiThread(new Runnable() {
            @Override
            public void run() {
                Log.d("SocketMsg: ", args[0].toString());
                //mSocket.emit("accessKey", accessKey);
                Toast.makeText(MainActivity.this.getApplicationContext(),
                        "Connect", Toast.LENGTH_LONG).show();
                for (Object o : args) {
                    Log.i("IO " + Socket.EVENT_CONNECT_ERROR, o.toString());
                }
            }
        });
    }
};
private Emitter.Listener onConnectError = new Emitter.Listener() {
    @Override
    public void call(final Object... args) {
        MainActivity.this.runOnUiThread(new Runnable() {
            @Override
            public void run() {
                Log.d("SocketMsg: ", args[0].toString());
                Toast.makeText(MainActivity.this.getApplicationContext(),
                        "Error", Toast.LENGTH_LONG).show();
                for (Object o : args) {
                    Log.i("IO " + Socket.EVENT_CONNECT_ERROR, o.toString());
                }
            }
        });
    }
};

它给予 'xhr poll error'

Answer 1

我建议你使用 socket.io.

在 Socket.io Blog

处有一个很好的客户端和服务器示例

Answer 2

现在使用以下技巧成功连接到 SockJS :)。

1. Place a web view and set its visibility to gone.
2. Use SockJS client in webview.
3. Create a javascript Interface that send data back to native code when get data from socket.
4. We have to call the function only once from native code.

JavaScript 界面：

public class WebAppInterface {
    Context mContext;

    /**
     * Instantiate the interface and set the context
     */
    WebAppInterface(Context c) {
        mContext = c;
    }

    /**
     * Show a toast from the web page
     */
    @JavascriptInterface
    public void msg(final String toast) {
        Log.d("msg: ", toast);
        if (toast.contains("{"))
            createMarkerList(toast);
    }
}

Webview 上的 SockJS 客户端：

<script type="text/javascript">
    function connectToSocket(url, accessKey){
        Android.msg(String("called"));
        var sock = new SockJS("http://api.example.in:8182/realtime");
        sock.onopen = function() {
            Android.msg("open");
            sock.send(JSON.stringify({
              key: accessKey
            }));
        };

        sock.onclose = function() {
            Android.msg("close");
        };

        sock.onmessage = function(e) {
            result = e.data.replace(/,\s*$/, '');
            Android.msg(String(result));
        };

        sock.onerror = function(e) {
            Android.msg("error");
        };
    }
</script>

调用js函数：

webView.loadUrl(javascript:connectToSocket(url, accessKey));

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