【发布时间】:2016-10-10 08:23:11
【问题描述】:
所以我有以下情景:
我有一个浏览器作为我的推送服务器, 我的移动应用作为客户端。
现在我想将浏览器与手机配对,并让它们都连接到由浏览器动态组成的同一个频道,这将向我的设备发送一条消息并告诉它连接到 X 频道。现在我被困在应用程序部分,了解如何让设备连接到通过 Pusher 发送的通道。
我在 JAVA 中有以下代码:
public void openPusherListener() {
//default settings for pusher
PusherOptions options = new PusherOptions().setCluster("eu");
final Pusher pusher = new Pusher(API_KEY, options);
pusher.connect(new ConnectionEventListener() {
@Override
public void onConnectionStateChange(ConnectionStateChange change) {
String socketId = pusher.getConnection().getSocketId();
Log.d("socketID", socketId);
}
@Override
public void onError(String message, String code, Exception e) {
System.out.println("There was a problem connecting!");
}
}, ConnectionState.CONNECTED);
Channel channel = pusher.subscribe("waitlist");//Default channel to join
channel.bind("test", new SubscriptionEventListener() {
@Override
public void onEvent(String channelName, String eventName, final String data) {
Gson gson = new Gson();
JsonObject mydata = gson.fromJson(data, JsonObject.class);
Log.d("PUSHER - channel", String.valueOf(mydata.get("channel")));
String myChannel = String.valueOf(mydata.get("channel")).replace('"', ' ');
Log.d("TEST MYCHANNEL", String.valueOf(mydata.get("channel")));
Channel channel = pusher.subscribe(myChannel);//This doesnt work.
channel.bind("test2", new SubscriptionEventListener() {
@Override
public void onEvent(String channelName, String eventName, final String data) {
Gson gson = new Gson();
JsonObject mydata = gson.fromJson(data, JsonObject.class);
Log.d("TEST2 PUSHER", String.valueOf(mydata.get("joinChannel")));
}
});
}
});
pusher.connect();
}
不幸的是,加入第二个频道不起作用,我收到以下错误:
I/System.out: There was a problem connecting!
在 Pusher 上我可以看到以下内容:
Invalid channel name ' test2 ' (bad chars)
现在我认为我必须找到一种方法来连接到频道而不向它发送报价,所以我想知道如何实现这一点?
【问题讨论】: