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java - Applet 和 Servlet 通信

转载 作者:行者123 更新时间:2023-12-01 15:33:35 24 4
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我有一个小程序需要向 servlet 提交分数,但它无法正常工作。

这是小程序的代码

private URLConnection getConnection() throws MalformedURLException, IOException {
URL serverAddress = null;
URLConnection conn = null;
serverAddress = new URL("http://localhost/GamesPortal/submitScore");
conn = serverAddress.openConnection();
conn.setDoOutput(true);
conn.setRequestProperty("Content-Type", "application/x-java-serialized-object");
return conn;
}

private void sendRecievedata(GameInfo info) {
try {
URLConnection c = this.getConnection();
OutputStream os = c.getOutputStream();
ObjectOutputStream oos = new ObjectOutputStream(os);
oos.writeObject(info);
oos.close();
} catch (Exception ex) {
ex.printStackTrace();
}
}

这是 servlet 代码

    try {
HttpSession s = request.getSession(true);

response.setContentType("application/x-java-serialized-object");
InputStream in = request.getInputStream();
ObjectInputStream ois = new ObjectInputStream(in);
GameInfo info = (GameInfo) ois.readObject();

if (info.getUserId() > 0) {
Scores score = new Scores();
score.submitScore(info);
}
} catch (Exception ex) {
ex.printStackTrace();
} finally {
}
response.setContentType("text/html;charset=UTF-8");
PrintWriter out = response.getWriter();
try {
out.println("<html>");
out.println("<head>");
out.println("<title>Servlet submitScore</title>");
out.println("</head>");
out.println("<body>");
out.println("<h1>Servlet submitScore at " + request.getContextPath() + "</h1>");
out.println("</body>");
out.println("</html>");
} catch {
ex.printStackTrace();
} finally {
out.close();
}

现在我尝试通过浏览器访问servlet,只是为了确保地址正确(确实如此),但由于某种原因,当我尝试从小程序本身访问它时,它无法连接。 (调试器甚至没有启动)。

(根据建议,将 ex.printStackTrace(); 添加到每个 try catch 中,但我不知道应该在什么地方或在哪里寻找它)

调用小程序的代码类似于以下内容: http://roseindia.net/jsp/simple-jsp-example/applet-in-jsp.shtml

<jsp:plugin code="Pong.class" name="Games/Pong/Pong" type="applet" width="800" height="600">
<jsp:params>
<jsp:param name="userId" value="<%= user.getUserId()%>" ></jsp:param>
</jsp:params>
</jsp:plugin>

这里有什么我忽略的东西吗?

最佳答案

我已经设法让它发挥作用。

这是小程序的代码:

    private URLConnection getServletConnection()
throws MalformedURLException, IOException {
URL urlServlet = new URL("http://localhost:8080/GamePortal/submitScore");
URLConnection con = urlServlet.openConnection();
con.setDoOutput(true);
con.setRequestProperty(
"Content-Type",
"application/x-java-serialized-object");
return con;

}

private void onSendData(GameInfo info) {

try {
// send data to the servlet
URLConnection con = getServletConnection();
OutputStream outstream = con.getOutputStream();
ObjectOutputStream oos = new ObjectOutputStream(outstream);
oos.writeObject(info);
oos.flush();
oos.close();
// receive result from servlet
InputStream instr = con.getInputStream();
ObjectInputStream inputFromServlet = new ObjectInputStream(instr);
String result = (String) inputFromServlet.readObject();
//JOptionPane.showMessageDialog(null, result);
inputFromServlet.close();
instr.close();
} catch (Exception ex) {
ex.printStackTrace();
}
}

这是 servlet 的代码:

protected void processRequest(HttpServletRequest request, HttpServletResponse response)
throws ServletException, IOException {
try {
response.setContentType("application/x-java-serialized-object");
// read a String-object from applet
// instead of a String-object, you can transmit any object, which
// is known to the servlet and to the applet
InputStream in = request.getInputStream();
ObjectInputStream inputFromApplet = new ObjectInputStream(in);
GameInfo score = (GameInfo) inputFromApplet.readObject();
System.out.println(score.getScore());

GameInfo info = score;

if (info.getUserId() > 0) {
Scores instance = new Scores();
instance.submitScore(info);
}

OutputStream outstr = response.getOutputStream();
ObjectOutputStream oos = new ObjectOutputStream(outstr);
oos.writeObject("reply");
oos.flush();
oos.close();
} catch (ClassNotFoundException ex) {
}
}

感谢您的帮助,请原谅我花了太长时间才回复。

关于java - Applet 和 Servlet 通信,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/9247062/

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