java - NetworkManager 受 AP 支持

Question

我想使用一项服务。我 100% 确信该服务可以正常工作。

服务电话

public void add(User user) {
ConnectionRequest con = new ConnectionRequest();

String url="http://localhost/NY/untitled/web/app_dev.php/user/new"
+ "?Fonctionuser="+user.getUserFunction()
+"&Fullname="+user.getUserName()
+"&Imageproduit="+user.getUserImage()
+"&Latitude="+user.getLatitude()
+"&Longitude="+user.getLongitude()
+"&State="+user.getUserState();
System.out.println(user.getUserState());

con.setUrl(url);
NetworkManager.getInstance().addToQueue(con); 
}

错误

Exception in thread "main" java.lang.NullPointerException
at com.codename1.io.NetworkManager.isAPSupported(NetworkManager.java:866)
at com.codename1.io.ConnectionRequest.<init>(ConnectionRequest.java:330)
at Services.UserServices.add(UserServices.java:18)
at Services.Main.main(Main.java:30)

主要方法

public static void main(String[] args) { 
  User user = new User(); 
  user.setLatitude(111); 
  user.setLongitude(111); 
  user.setUserFunction( UserFunction.Client.ordinal()); 
  user.setUserImage("uezfniez"); 
  user.setUserState((int)UserState.Hold.ordinal()); 
  user.setUserName("jamel"); 
  UserServices userServices = new UserServices(); 
  userServices.add(user); 
} 

Answer 1

看来您需要在操作监听器中编写代码

registerButton.addActionListener(e -> {
        con.setUrl("http://localhost/NY/untitled/web/app_dev.php/user/"
                + "new?Fonctionuser=2&Fullname=jamel&Imageproduit=uezfniez&Latitude=111&Longitude=111&State=2");
        NetworkManager.getInstance().addToQueue(con);
    });