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我在 java 中创建了一个 web 服务，它应该返回我电脑中文件夹的内容。我想用文件名填写一个列表，当用户点击一个时，它将被我的应用程序打开。对于测试，我只想至少获得文件名。我使用了以下代码：

简单网络服务：

@WebService
@SOAPBinding(style=Style.RPC)
public interface SimpleService {
   @WebMethod
   File getFiles();
}

SimpleWebServiceImpl：

@WebService(endpointInterface="com.medex.webServiceMegXsoft.SimpleService")
public class SimpleServiceImpl implements SimpleService{
    public File getFiles() {
        File directory = new 
            File("C:\\Users\\student\\Desktop\\MegXsoftMobile\\");
        return directory;
    }
}

简单服务发布者：

public class SimpleServicePublisher {
    public static void main(String[] args) {
        Endpoint.publish("http://192.168.0.58:9000/simple",
                new SimpleServiceImpl());
    }
}

我的接收器：

public class Receiver {

private final String NAME_SPACE = "http://192.168.0.71:9000/";
private final String URL = "http://192.168.0.71:9000/simple?wsdl";
private final String SOAP_ACTION = "\"getFiles\"";
private final String METHOD_NAME = "getFiles";
private Object resultsRequestSOAP;

public File getFilesFromXML() {
    SoapObject request = new SoapObject(NAME_SPACE, METHOD_NAME);
    SoapSerializationEnvelope envelope = new SoapSerializationEnvelope(
            SoapEnvelope.VER11);

    envelope.setOutputSoapObject(request);
    AndroidHttpTransport androidHttpTransport = new AndroidHttpTransport(URL);
    try
    {
        androidHttpTransport.call(SOAP_ACTION, envelope);
        resultsRequestSOAP =  envelope.getResponse();
    } catch (SoapFault e) {
        e.printStackTrace();
    }catch (XmlPullParserException e){
        e.printStackTrace();
    }catch (IOException e) {
        e.printStackTrace();
    }catch (Exception e){
        e.printStackTrace();
    }

    return (File)resultsRequestSOAP;
}

}

最后是活动：

ArrayList<String> stringTable;
File files = null;
ProgressDialog dialog;
Thread thread;
Context context;

@Override
public void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    context = this;
    dialog = ProgressDialog.show(this, null, "loading", true, false);

    thread = new Thread(new Runnable() {
        public void run() {
            Receiver receiver = new Receiver();
            files = receiver.getFilesFromXML();
            dialog.dismiss();
        }
    });
    thread.start();

    final Handler h = new Handler();
    h.postDelayed(new Runnable() {

        public void run() {
            if (thread.getState() == Thread.State.TERMINATED) {
                int i = 0;

                stringTable = new ArrayList<String>();

                for (File file : files.listFiles()) {
                    stringTable.add(i, file.getName());
                    i++;
                }
                final ArrayAdapter<String> adapter = new ArrayAdapter<String>(
                        context, R.layout.item, stringTable);
                ListView listView = getListView();
                listView.setAdapter(adapter);

                listView.setOnItemClickListener(new OnItemClickListener() {

                    public void onItemClick(AdapterView<?> parent,
                            View view, int position, long id) {
                        System.out.println(stringTable.get(position));
                    }
                });

            }
            h.postDelayed(this, 1000);
        }
    }, 1000);
}

它似乎发布了一些东西，就像我去的时候我http://192.168.0.58:9000/simple?wsdl在 XML 中有一些行。但我觉得我无法File在我的 android 中收到。我收到了这个错误：

06-20 13:55:55.759：W/System.err(13370)：org.xmlpull.v1.XmlPullParserException：预期：END_TAG {http://schemas.xmlsoap.org/soap/envelope/}正文（位置：END_TAG @1:290 在 java.io.InputStreamReader@4053a7f0)

如果有人知道如何更轻松地获取文件列表，欢迎您:) 我错在哪里？

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1 回答 1

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嗯，不确定，但我不相信有一种真正的方法来支持 SOAP 中的“文件”构造，但您可能能够构造带有 mime 编码附件的消息。这个关于 PHP 服务器的类似问题的链接导致This description on how to do it in Java and Jax/RPC

祝你好运！

于 2012-06-20T14:56:10.517 回答