Question

我有一个wcf服务，根据输入返回true或false（检查用户登录）。我试图在Android应用程序中实现它，但每当我按下按钮时，应用程序崩溃。这是代码：

btnLogin.setOnClickListener(new OnClickListener() {

        @Override
        public void onClick(View v) {
            // TODO Auto-generated method stub

            email = editEmail.getText().toString();
            pass = editPass.getText().toString();

            if(email.matches("") || pass.matches("")){
                txtresult.setText("Please fill in the information!");
                return;
            }


            if(!isConnected){
                txtresult.setText("You don't have internet connection!");
                return;
            }

            new MyTask().execute();

            if(status.equals("true")){
                //savePreferences();
                finish();
            }
            else{
                txtresult.setText("Wrong login information!");
            }
        }
    });

AsyncTask：

public class MyTask extends AsyncTask<Void,Void,Void> {

    String email1 = editEmail.getText().toString();
    String pass1 = editPass.getText().toString();

    @Override
    protected Void doInBackground(Void... params) {
        HttpClient httpclient = new DefaultHttpClient();
        HttpGet httpGet = new HttpGet("http://wswob.somee.com/wobservice.svc/checkLogin/"+email1+"/"+pass1);
        HttpContext localContext = new BasicHttpContext();

        HttpResponse response;

          try {

            response = httpclient.execute(httpGet, localContext);
            HttpEntity entity = response.getEntity();
            status = EntityUtils.toString(entity);



          } catch (ClientProtocolException e) {
            // TODO Auto-generated catch block
            e.printStackTrace();
          } catch (IOException e) {
            // TODO Auto-generated catch block
            e.printStackTrace();
          }

        return null;
    }
}

这工作正常并且突然停止工作......由于我2周的经验，我不确定如何调试。有什么想法吗？

Answer 1

这一行：

if(status.equals("true")){

崩溃，因为status尚未分配，抛出NullPointerException。

根据定义，AsyncTasks是异步的。这意味着一旦进行异步调用，代码控件将立即返回并在另一个线程中启动异步工作。

执行此操作的正确方法是使用AsyncTask回调onPostExecute()，并在那里检查您的状态。

public class MyTask extends AsyncTask<Void,Void,Void> {

    String email1 = editEmail.getText().toString();
    String pass1 = editPass.getText().toString();

    @Override
    protected Void doInBackground(Void... params) {
        HttpClient httpclient = new DefaultHttpClient();
        HttpGet httpGet = new HttpGet("http://wswob.somee.com/wobservice.svc/checkLogin/"+email1+"/"+pass1);
        HttpContext localContext = new BasicHttpContext();

        HttpResponse response;

          try {

            response = httpclient.execute(httpGet, localContext);
            HttpEntity entity = response.getEntity();
            status = EntityUtils.toString(entity);



          } catch (ClientProtocolException e) {
            // TODO Auto-generated catch block
            e.printStackTrace();
          } catch (IOException e) {
            // TODO Auto-generated catch block
            e.printStackTrace();
          }

        return null;
    }

    @Override
    protected Void onPostExecute(Void result){
        super.onPostExecute(result);
        //TODO: your code here. check your status and handle from there
    }
}

调用AsyncTask Android时崩溃

1 个答案: