Question

我错过了一些非常关键的东西，但我看不出是什么。有人可以帮助。我可能错过了一些非常愚蠢的东西，但我无法启动我的onItemClick。

的onCreate ....

    @Override
public void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    Intent intent = getIntent();
    actu_ip = intent.getStringExtra(IPEntry.ACTUALSMARTIP);
    setContentView(R.layout.act_ipcontrol);

    mainListView = (ListView) findViewById( R.id.mainListView );
    String[] options = new String[] { "All in to 1", "Spare"};  

    ArrayList<String> optionsList = new ArrayList<String>();
    optionsList.addAll( Arrays.asList(options) );
    listAdapter = new ArrayAdapter<String>(this, R.layout.simplerow, optionsList);
    mainListView.setAdapter( listAdapter ); 

    try {
        Toast.makeText(IPControl.this, "Please wait...Connecting...", Toast.LENGTH_SHORT).show();
        new AsyncAction().execute();
    } catch(Exception e) {
        e.printStackTrace();
    }
}

private class AsyncAction extends AsyncTask<String, Void, String> {
    protected String doInBackground(String... args) { 
        try {
            InetAddress serverAddr = InetAddress.getByName(actu_ip);
            socket = new Socket(serverAddr, REDIRECTED_SERVERPORT);
            OutputStreamWriter osw = new OutputStreamWriter(socket.getOutputStream());
            BufferedWriter bw = new BufferedWriter(osw);
            out = new PrintWriter(bw, true); 
            in = new BufferedReader(new InputStreamReader(socket.getInputStream()));
            while (! in .ready());
            readBuffer();
            out.println("root\r\n");
            while (! in .ready());
            readBuffer();
            out.println("root\r\n");
            while (! in .ready());
            readBuffer();
            out.println("[verbose,off\r\n");
            while (! in .ready());
            String msg = "";
            while ( in .ready()) {
                msg = msg + (char) in .read();
            }
        } catch (UnknownHostException e) {
            e.printStackTrace();
        } catch (IOException e) {
            e.printStackTrace();
        }

        return null;//returns what you want to pass to the onPostExecute()
    }

    protected void onPostExecute(String result) {

        Toast.makeText(IPControl.this, "Connected", Toast.LENGTH_SHORT).show();

        //results the data returned from doInbackground

        IPControl.this.data = result;

    }
}

private String readBuffer() throws IOException {
    String msg = "";

    while(in.ready()) {
        msg = msg + (char)in.read();
    }
    //System.out.print(msg);
    if(msg.indexOf("SNX_COM> ") != -1) return msg.substring(0, msg.indexOf("SNX_COM> "));
    else if(msg.indexOf("SCX_COM> ") != -1) return msg.substring(0, msg.indexOf("SCX_COM> "));
    else return msg;
}
}

我想要发起什么......

        public void onItemClick(AdapterView<?> arg0, View arg1, int pos,
            long arg3) {

        try {
            new AsyncAction1().execute();
        } catch(Exception e) {
            e.printStackTrace();
        }
    }

    private class AsyncAction1 extends AsyncTask<String, Void, String> {
        protected String doInBackground(String... args) { 
            try {
                out.println("[c,l#,i1,o*\r\n");
                //System.out.print("root\r\n");
                while(! in .ready());


            } catch (UnknownHostException e) {
                e.printStackTrace();
            } catch (IOException e) {
                e.printStackTrace();
            }

            return null;//returns what you want to pass to the onPostExecute()
        }

        protected void onPostExecute(String result) {

            //results the data returned from doInbackground
            Toast.makeText(IPControl.this, "Command Sent", Toast.LENGTH_SHORT).show();

            IPControl.this.data = result;

        }
    }

Answer 1

我还没有在代码中看到listview的setOnItemClickListener方法。你实现了吗？试试以下

mainListView.setAdapter( listAdapter ); 
mainListView.setOnItemClickListener(new OnItemClickListener(){
   public void onItemClick(AdapterView<?> parent, View view, int position,
        long id) {

      try {
        new AsyncAction1().execute();
      }catch(Exception e) {
        e.printStackTrace();
   }
});

Answer 2

感谢你的帮助，我解决了我的问题。

我只是没有按照正确的顺序做事。

@Override
public void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    Intent intent = getIntent();
    actu_ip = intent.getStringExtra(IPEntry.ACTUALSMARTIP);
    setContentView(R.layout.act_ipcontrol);

    mainListView = (ListView) findViewById( R.id.mainListView );
    final String[] options = new String[] { "All in to 1", "Spare"};  

    ArrayList<String> optionsList = new ArrayList<String>();
    optionsList.addAll( Arrays.asList(options) );
    listAdapter = new ArrayAdapter<String>(this, R.layout.simplerow, optionsList);
    mainListView.setAdapter( listAdapter ); 
    mainListView.setOnItemClickListener(new OnItemClickListener(){

        public void onItemClick(AdapterView<?> arg0, View arg1, int pos, long arg3) {
            try {
                    if(pos == 0) {
                        AsyncAction1 a = new AsyncAction1();
                        a.setCmd("[c,l#,i1,o*\r\n");
                        a.execute();
                    }
                } catch(Exception e) {
                    e.printStackTrace();
                }
        }
    });

然后......

private class AsyncAction1 extends AsyncTask<String, Void, String> {
    String cmd;

    public void setCmd(String c) {
        cmd = c;
    }

    protected String doInBackground(String... args) {
        try {
            out.println(cmd);
            //System.out.print("root\r\n");
            while(! in .ready());
        } catch (UnknownHostException e) {
            e.printStackTrace();
        } catch (IOException e) {
            e.printStackTrace();
        }
        return null;//returns what you want to pass to the onPostExecute()
    }

    protected void onPostExecute(String result) {

        //results the data returned from doInbackground
        Toast.makeText(IPControl.this, "Command Sent", Toast.LENGTH_SHORT).show();

        IPControl.this.data = result;
    }
}
}

onItem单击时无法正常工作

2 个答案: