从Java应用程序调用Servlet

Question

我想从Java应用程序调用Servlet。 问题是，调用似乎没有到达Servlet。 我没有得到任何错误，但没有到达Servlet中的第一个输出“doPost”。 如果我在网络浏览器中打开URL，我当然得到了GET不支持的错误等，但至少我看到，有些事情发生了。

我使用以下代码（ActionPackage类只包含参数Vector并且是Serializable）：

Java应用程序：

    ActionPackage p = new ActionPackage();
    p.addParameter("TEST", "VALUE");

    System.out.println(p);

    URL gwtServlet = null;
    try {
        gwtServlet = new URL("http://localhost:8888/app/PushServlet");
        HttpURLConnection servletConnection = (HttpURLConnection) gwtServlet.openConnection();
        servletConnection.setRequestMethod("POST");
        servletConnection.setDoOutput(true);

        ObjectOutputStream objOut = new ObjectOutputStream(servletConnection.getOutputStream());
        objOut.writeObject(p);
        objOut.flush();
        objOut.close();

    } catch (MalformedURLException e) {
        // TODO Auto-generated catch block
        e.printStackTrace();
    } catch (IOException e) {
        // TODO Auto-generated catch block
        e.printStackTrace();
    }
}

Servlet的：

public class PushServlet extends HttpServlet {

public void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
    System.out.println("doPost");
    ObjectInputStream objIn = new ObjectInputStream(request.getInputStream());

    ActionPackage p = null;
    try {
        p = (ActionPackage) objIn.readObject();

    } catch (ClassNotFoundException e) {
        e.printStackTrace();
    }

    System.out.println("Servlet received p: "+p);       
}

}

出了什么问题？

谢谢。

Answer 1

只要你调用任何get方法， URLConnection就会被懒惰地执行。

将以下内容添加到代码中以实际执行HTTP请求并获取servlet响应主体。

InputStream response = servletConnection.getInputStream();

也可以看看：

• 如何使用java.net.URLConnection来触发和处理HTTP请求？

Answer 2

public void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
    try {
        System.out.println("doPost");
        ObjectInputStream objIn = new ObjectInputStream(request.getInputStream());
        ActionPackage p = null;
        p = (ActionPackage) objIn.readObject();
        System.out.println("Servlet rece p: "+p);       
    } catch (Throwable e) {
        e.printStackTrace(System.out);
    }
}

Answer 3

尝试在try / catch块中包装doPost的整个主体：

public void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
    try {
        System.out.println("doPost");
        ObjectInputStream objIn = new ObjectInputStream(request.getInputStream());
        ActionPackage p = null;
        p = (ActionPackage) objIn.readObject();
        System.out.println("Servlet received p: "+p);       
    } catch (Throwable e) {
        e.printStackTrace(System.out);
    }
}

然后再次查看Servlet输出日志文件或窗口以获取可能有用的新异常。