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JAVA-是否可以将接收到的servlet值传递给.java文件,但是在另一个项目中?

[英]JAVA - Is there possible pass a servlet value received to a .java file, but in another project?

good afternoon. 下午好。 I have a .JSP file that send a value to a servlet file. 我有一个.JSP文件,可将值发送到servlet文件。 I have another project, where there is a java file in that i want to get this servlet value received by .JSP file. 我有另一个项目,其中有一个Java文件,我想获取该.JSP文件接收的servlet值。

My question is if is there possible pass this servlet value to this .java file in another project ? 我的问题是,是否有可能将此servlet值传递给另一个项目中的.java文件? Or can i send by .JSP file directly ? 还是可以直接通过.JSP文件发送?

I created a Java Class in my Web Application, but now, how can i pass the servlet parameter to this java class ? 我在Web应用程序中创建了一个Java类,但是现在,如何将servlet参数传递给该Java类?

I will put the code down here: 我将代码放在这里:

Servlet code: Servlet代码:

public class ServletJava extends HttpServlet {

    protected void processRequest(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
     response.setContentType("text/html;charset=UTF-8");

     /*I WANT TO INSTANCE THE PARAMETER HERE TO SEND TO OTHER CLASS*/
    }

    @Override
    protected void doGet(HttpServletRequest request, HttpServletResponse response)
            throws ServletException, IOException {
        processRequest(request, response);
    }

    @Override
    protected void doPost(HttpServletRequest request, HttpServletResponse response)
            throws ServletException, IOException {
        processRequest(request, response);
    }

    @Override
    public String getServletInfo() {
        return "Short description";
    }

}

The Class java: 类java:

public class RecebeServlet {
     public static void main(String[] args) throws Exception {
      /*I WANT RECEIVE THE PARAMETER HERE !! */
   }
}

It is probably bad practice to have a class with only one main mathod. 上一堂课只有一个主要方法可能是不好的做法。 I am sure that you can divide the logic from that class into several methods instead of having one giant main . 我敢肯定,您可以将该类的逻辑分为几种方法,而不用拥有一个庞大的main

But anyway, since you asked: Just call your main method as you would call any static method: 但是无论如何,既然您问过:就像调用任何静态方法一样,只需调用main方法即可:

String[] args = new String[1];
args[0] = yourValue;
RecebeServlet.main(args);

But again, please consider refactoring your RecebeServlet class. 但同样,请考虑重构您的RecebeServlet类。 You do not want to have program logic in your main method. 您不想在主方法中包含程序逻辑。

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