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我编写了这个小应用程序，它从 Web 服务中获取一些数据并将它们放入自定义 ListView：

肥皂电话

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;
}

异步任务

 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();
    }

然后我获取字符串变量，拆分我的数据并将它们放入自定义列表中......

它工作得很好，但问题是速度很慢所以我尝试创建一个 ProgressDialog 但有一些问题......

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();
    }

}

进度图像不动，它似乎被冻结了，如果我把它放在 LineCall asyncTask 中，则不会显示 ProgressDialog...

我在谷歌发现这是由于 string = line.get(); 这需要很长时间才能运行...

那么当 get() 工作时如何放置 ProgressDialog 呢？

4

1 回答 1

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如果你调用AsyncTask.get()，它将阻塞主线程直到doInBackground完成。因此，您不应该调用AsyncTask.get()，而是应该处理结果并在AsyncTask.onPostExecute()方法中执行 UI 操作。这是 的设计行为AsyncTask。

在您的 StartDeapartures asynctask 中，为什么要在 doInBackground 中启动活动？使用 AsyncTask 启动活动有什么意义？

于 2013-11-03T20:21:06.207 回答