1

我有一个简单的 android 应用程序来访问 Web 服务。我想显示一个对话框"Loading . . ."，直到加载结果并在加载结果后关闭。我已经使用了这段代码，但这没有显示加载消息：

public class LoadingActivity extends Activity {
   private static final String SOAP_ACTION = "http://ws.eretailer.com/";
   private static final String METHOD_NAME = "dataSender";
   private static final String NAMESPACE = "http://ws.eretailer.com/dataSender/";
   private static final String URL = "http://175.157.128.207:8085/Eretailer/services/EretailerService?wsdl";
/** Called when the activity is first created. */
@Override
public void onCreate(Bundle savedInstanceState){
    super.onCreate(savedInstanceState);
    setContentView(R.layout.main);
    final ProgressDialog pd = new ProgressDialog(this);
    pd.setProgressStyle(ProgressDialog.STYLE_HORIZONTAL);
    pd.setMessage("Loading. . .");
    pd.show();
    SoapObject request = new SoapObject(NAMESPACE, METHOD_NAME);          

    SoapSerializationEnvelope envelope = new SoapSerializationEnvelope(SoapEnvelope.VER11);

    envelope.setOutputSoapObject(request);
    pd.show();
    HttpTransportSE ht = new HttpTransportSE(URL);
    try {

        ht.call(SOAP_ACTION, envelope);
        SoapPrimitive response = (SoapPrimitive)envelope.getResponse();
        SoapPrimitive s = response;

        pd.dismiss();
           String str = s.toString();
           String arr1[] = str.split(" ");

           TextView tv = new TextView(this);

           for(int i = 0; i<arr1.length;i++){
           tv.append("order ID :"+arr1[i]+"\n");
          }
           setContentView(tv);

       } catch (Exception e) {

           e.printStackTrace();
       }
}
}

我怎样才能纠正这个问题？

4

3 回答 3

2

嗨试试下面的代码

 private class DownloadWebPageTask extends AsyncTask<String, Void, String> {
    @Override
    protected void onPreExecute() {
        pd.show();
        super.onPreExecute();
    }
    @Override
    protected String doInBackground(String... urls) {
        String response = "";
         HttpTransportSE ht = new HttpTransportSE(URL);
            try {

                ht.call(SOAP_ACTION, envelope);
                SoapPrimitive response = (SoapPrimitive)envelope.getResponse();
                SoapPrimitive s = response;

                pd.dismiss();
                   String str = s.toString();
                   String arr1[] = str.split(" ");

                   TextView tv = new TextView(this);

                   for(int i = 0; i<arr1.length;i++){
                   tv.append("order ID :"+arr1[i]+"\n");
                  }
                   setContentView(tv);

               } catch (Exception e) {

                   e.printStackTrace();
               }
        return response;
    }

    @Override
    protected void onPostExecute(String result) {
        pd.dismiss();
    }
}

资源：http ://www.vogella.com/articles/AndroidPerformance/article.html

于 2012-05-05T08:35:08.890 回答
1

最好将Android-AsyncTask用于您的 Web 请求。并从AsyncTask 开始并ProgressDialog在其中关闭它（在此处更新您的 UI 以获取 Web 响应的结果）。还将Web-Request 代码放入AsyncTask..onPreExecute()onPostExecute()doInBackGround()

例如看Android AsyncTask Example。

这永远不会阻塞您的Main-UI Thread。以及更好的用户体验性能。

于 2012-05-05T08:22:26.830 回答
0

首先评论 pd.dismiss() 并检查它是否正在加载，然后在下载完成后使用 pd.dismiss() 。

于 2012-05-05T08:16:03.420 回答