java - Java Servlet 中参数未正确传递

Question

我正在尝试编写一个简单的 java servlet 来列出目录中的文件。该路径存储在 web.xml 的 init-param 中。当我调用 getInitParameters() 时，它返回目录路径。但是当我尝试将它传递给处理程序时，它返回 null。不确定我做错了什么。有什么帮助吗？

import java.io.File;
import java.io.IOException;
import java.io.PrintWriter;

import javax.servlet.ServletException;

import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;

interface Handler {
  public void doGet (HttpServletRequest request, HttpServletResponse response) 
    throws IOException; 
}

class DispatchChoice {
  public final String param; 
  public final GetHandler getHandler; 
  public DispatchChoice (String param, Handler getHandler) 
  {
    this.param = param;
    this.getHandler = getHandler;
  }
}

public class MyServlet extends HttpServlet
{
    String value;
    public void init() throws ServletException {
        value = getInitParameter("addressfile"); // correct value is saved here
        System.out.println("Init value : "+value);
    }
  DispatchChoice myChoice = new DispatchChoice("/test1", new FileHandler(value));

  public void doGet (HttpServletRequest request, HttpServletResponse response) 
    throws IOException
  {
        myChoice.getHandler.doGet(request, response);
  }
}

class FileHandler implements Handler {
    private String place;
    public FileHandler (String value){
        this.place = value; // this is NULL, not the value from above
        System.out.println("Param value : " + value);
    }

    public void doGet(HttpServletRequest request, HttpServletResponse response)
            throws IOException {
        File directory = new File(place); //is NULL
        File[] files = directory.listFiles();
        PrintWriter pw = response.getWriter();
        for (int index = 0; index < files.length; index++) {
            pw.println(files[index].getName());
        }
    }
}

web.xml

<servlet>
<servlet-name>ListManagerServlet</servlet-name>
<servlet-class>savva.listmanagerservlet.ListManagerServlet</servlet-class>
<init-param>
    <param-name>addressfile</param-name>
    <param-value>d:\\temp\\</param-value>
</init-param>
</servlet>
<servlet-mapping>
<servlet-name>ListManagerServlet</servlet-name>
<url-pattern>/ListManagerServlet</url-pattern>
</servlet-mapping>

Answer 1

DispatchChoice myChoice = new DispatchChoice("/test1", new FileHandler(value));

该行在init()之前执行，因此value仍然是null并且尚未分配!相反，将赋值移到 init() 内，如下所示:

DispatchChoice myChoice;

public void init() throws ServletException
{
    value = getInitParameter("addressfile"); // correct value is saved here
    myChoice = new DispatchChoice("/test1", new FileHandler(value));
    System.out.println("Init value : "+value);
}