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[英]Which is the correct format to return a WebService in Java for Android
[英]How to return a File in a WebService for Android in Java
我已经在Java中制作了一个Web服务,该服务应该返回我PC中文件夹的内容。 我想用文件名填充一个列表,当用户点击一个列表时,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。 我使用以下代码:
SimpleWebService:
@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;
}
}
SimpleServicePublisher:
public class SimpleServicePublisher {
public static void main(String[] args) {
Endpoint.publish("http://192.168.0.58:9000/simple",
new SimpleServiceImpl());
}
}
而我的接收者:
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;
}
}
最后的活动:
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);
}
似乎发布了一些内容,因为当我转到http://192.168.0.58:9000/simple?wsdl
我在XML中有几行内容。 但是我感觉我无法在Android中接收File
。 我收到此错误:
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 java.io.InputStreamReader@4053a7f0中的@ 1:290)
如果有人对如何简化文件列表有任何想法,欢迎您:)我在哪里错了?
嗯,不确定,但是我不相信有一种真正的方法来支持SOAP中的“文件”构造,但是您也许能够构造带有mime编码附件的消息。 此链接在PHP服务器上引发了类似问题,导致此链接 描述了如何在Java和Jax / RPC中进行操作
祝好运!
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