java webstart问题怎么解决

这篇文章主要介绍“java webstart问题怎么解决”，在日常操作中，相信很多人在java webstart问题怎么解决问题上存在疑惑，小编查阅了各式资料，整理出简单好用的操作方法，希望对大家解答”java webstart问题怎么解决”的疑惑有所帮助！接下来，请跟着小编一起来学习吧！

当时碰到的几个技术问题是：

1.从web传递相关的参数给application,

　　解决办法：用动态jnlp文件（jsp实现jnlp）,同时用到如下传参办法　　　

application-desc Element

The application element indicates that the JNLP file is launching an application (as opposed to an applet). The application element has an optional attribute, main-class, which can be used to specify the name of the application's main class, i.e., the class that contains the public static void main(String argv[]) method where execution must begin.

The main-class attribute can be omitted if the first JAR file specified in the JNLP file contains a manifest file containing the main class.

Arguments can be specified to the application by including one or more nested argument elements. For example:

<application-desc main-class="Main">
<argument>arg1argument>
<argument>arg2argument>
application-desc>

2.如何将application处理的结果传回给web server

解决办法，用URLConnection结合从jnlp中传来的web url (为一个后台处理的servlet地址),sessionID（用于识别当前用户，权限等判断）去创建一个新的url对象，并通过它在application和web server之间传递数据。在后台的servlet中通过sessionid,从session listener中找到当前用户，

　　private String getStringPostRequest(String command) throws Exception {
DataOutputStream dos=null;
ObjectInputStream dis=null;
try {
URLConnection urlConn = new URL(webServerStr).openConnection();
urlConn.setDoOutput(true);
urlConn.setDoInput(true);
urlConn.setAllowUserInteraction(false);
urlConn.setUseCaches(false);
urlConn.setRequestProperty(
"Content-Type",
"application/x-www-form-urlencoded");

dos = new DataOutputStream(urlConn.getOutputStream());
dos.writeBytes(command + "&sessionId=" + this.sessionId);
dos.close();
// read input from servlet
dis =
new ObjectInputStream(urlConn.getInputStream());
String ret = dis.readObject().toString();
dis.close();
return ret;
} catch (Exception e) {
throw e;
} finally{
if ( dos!=null) dos.close();
if ( dis!=null) dis.close();
}
}

后台sevlet:

public void doGet(HttpServletRequest request, HttpServletResponse response)
throws IOException, ServletException
{
HttpSession hSession = request.getSession();
System.out.println("Application:" + hSession.getId());
if(MyListener.getSessionById(request.getParameter("sessionId")) != null)
hSession = MyListener.getSessionById(request.getParameter("sessionId"));
System.out.println("OK" + hSession);

..............}

sessionlistener:

import java.util.HashMap;
import java.util.Map;
import javax.servlet.ServletContextEvent;
import javax.servlet.ServletContextListener;
import javax.servlet.http.*;

public class SessionsListener
implements ServletContextListener, HttpSessionListener
{

static Map map = new HashMap();

public SessionsListener()
{
}

public void contextInitialized(ServletContextEvent servletcontextevent)
{
}

public void contextDestroyed(ServletContextEvent servletcontextevent)
{
}

public void sessionCreated(HttpSessionEvent httpsessionevent)
{
HttpSession httpsession = httpsessionevent.getSession();
map.put(httpsession.getId(), httpsession);
}

public void sessionDestroyed(HttpSessionEvent httpsessionevent)
{
HttpSession httpsession = httpsessionevent.getSession();
map.remove(httpsession.getId());
}

public static HttpSession getSessionById(String s)
{
return (HttpSession)map.get(s);
}

}

3.jar包数字签名问题

4.java webstart cache问题即:JNLP file caching

http://forum.java.sun.com/thread.jspa?forumID=38&threadID=556847

(1)

If you remove the href= parameter from the jnlp tag, Java Web Start 1.4.2 will not cache the jnlp file.
1.5.0 still will, but if you also remove the

(2)

It seems the issue is with generated JNLP files.

Try the following:

response.addDateHeader("Date", Calendar.getInstance().getTime().getTime());
response.addDateHeader("Last-Modified", Calendar.getInstance().getTime().getTime());

Seems to have solved the problem for us.

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