CODEWITHSUNDEEP
javaandroidspringrestspring-mvc
Mehdi Akbarian Rastaghi
Mehdi Akbarian Rastaghi

Reputation: 467

Send Data to Server with Spring api's

i want to send data to my Restful Spring web service from my android client by Spring api's but i got below exeption ca anyone help me? for example i want to register a user to my server,on my server side i have register method like this:

@RequestMapping(value = "/register")
public boolean registerNewUser(@RequestParam(name = "user")User user){
    if(user!=null){
        try {
            SessionFactory sessionFactory=HibenateConnection.connectingHibernate();
            Session session=sessionFactory.openSession();
            session.beginTransaction();
            session.save(user);
            session.getTransaction().commit();
            session.close();
            return true;
        }catch (Exception e){
            e.printStackTrace();
        }
    }
    return false;
}

and on my client side i open connection in Asynctask Class like this:

@Override
protected ResponseHandler doInBackground(Object... params) {
    try {
        serverResponse=new ServerResponse();
        result = restTemplate.getForObject(URL+conditions,tClass,params);
        serverResponse.singleResult=result;
        serverResponse.isReady=true;
        responseHandler.onServerRespond(serverResponse);
    }catch (Exception e){
        serverResponse.isReady=false;
        serverResponse.message=e.getMessage();
    }
    return responseHandler;
}

and i called this method on my Activity with my user:

public static void registerUser(User user,ResponseHandler responseHandler){
    String condition="/user_api/register?user=";
    new Core<Long>(condition,Long.class,responseHandler).execute(user);
}

i got Exeption at line restTemplate.getForObject(URL+conditions,tClass,params); in my Core AsynTask Class and the exeption is:

Could not create URI object: Illegal character in scheme at index 0: 192.168.1.2:/3000/user_api/register?user

my server Url is 192.168.1.2:3000

Upvotes: 0

Views: 144

Answers (1)

varren
varren

Reputation: 14731

Answer is simple, but usually hard to debug, because it is so easy to forget to add protocol part http:// before local url 192.168.1.2. So end url in your case should look like this:

URL="http://192.168.1.2:3000"

Upvotes: 1

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