Question

我写了这个小应用程序，从Web服务中获取一些数据并将它们放在自定义ListView中：

SOAP CALL

public String getVoy(String voy) {
    String SOAP_ACTION = "XXXXXXXXXX";
    String METHOD_NAME = "XXXXXXXXXX";
    String NAMESPACE = "urn:DefaultNamespace";
    String URL = "http://xxxxxxxxxxxx/xxxxx.nsf/xxxxxxx?wsdl";

    SoapObject request = new SoapObject(NAMESPACE, METHOD_NAME);
    request.addProperty("VOY", voy);
    SoapSerializationEnvelope envelope = new SoapSerializationEnvelope(SoapEnvelope.VER11);
    envelope.setOutputSoapObject(request);
    HttpTransportSE androidHttpTransport = new HttpTransportSE(URL);
    try {
        androidHttpTransport.call(SOAP_ACTION, envelope);
        final SoapObject result = (SoapObject) envelope.bodyIn;
        rt = (result != null) ? result.getProperty(0).toString() : "Nessuna risposta 1";

    } catch (IOException e) {
        // TODO Auto-generated catch block
        e.printStackTrace();
    } catch (XmlPullParserException e) {
        // TODO Auto-generated catch block
        e.printStackTrace();
    } // Invio il webservice

    return rt;
}

的AsyncTask

 private class LineCall extends AsyncTask<String, Void, String> {

    @Override
    protected String doInBackground(String... params) {
        final String text = WebService.getInstance().getVoy(params[0]);

        return text;
    }
}

调用AsyncTask并获取数据

LineCall line = new LineCall();
    line.execute(voy);

    try {
        **string** = line.get();
    } catch (InterruptedException e) {
        // TODO Auto-generated catch block
        e.printStackTrace();
    } catch (ExecutionException e) {
        // TODO Auto-generated catch block
        e.printStackTrace();
    }

然后我使用字符串变量，拆分我的数据并将它们放入自定义列表...

效果很好，但问题是非常慢所以我试图创建一个ProgressDialog，但是有一些问题......

如果我把它放在上一个活动中：

 Button partenze = (Button) findViewById(R.id.partenze);
    partenze.setOnClickListener(new OnClickListener() {

        @Override
        public void onClick(View v) {

            StartDeapartures call = new StartDeapartures();
            call.execute();

        }
    });
}
private class StartDeapartures extends AsyncTask<Void, Void, Void> {

    private ProgressDialog dialog;

    @Override
    protected void onPreExecute() {
        // TODO Auto-generated method stub
        super.onPreExecute();
        dialog = new ProgressDialog(Home.this);
        dialog.setTitle("Loading..");
        dialog.setCancelable(false);
        dialog.show();
    }

    @Override
    protected Void doInBackground(Void... params) {
        Intent intent = new Intent(getApplicationContext(), Partenze.class);
        startActivity(intent);
        return null;
    }

    @Override
    protected void onPostExecute(Void result) {
        // TODO Auto-generated method stub
        super.onPostExecute(result);
        dialog.dismiss();
    }

}

进度图像不移动，似乎冻结了，如果我把它放在LineCall asyncTask中，则不会显示ProgressDialog ......

我在google中发现这是du-string = line.get（）;这需要很长时间才能运行......

那么当get（）工作时我怎么能放一个ProgressDialog？

Answer 1

如果您致电AsyncTask.get()，它将阻止主线程，直到doInBackground完成。因此，您不应该调用AsyncTask.get()，而应该处理结果并在AsyncTask.onPostExecute()方法中执行UI操作。这是AsyncTask的设计行为。

在StartDeapartures asynctask中，为什么要在doInBackground中启动一个活动？使用AsyncTask启动活动有什么意义？

调用AsyncTask.get（）时的ProgressBar

1 个答案: