对HttpServlet的扩展与设计

这几天在学习Extjs中,需要用到后台与前台作数据交互,所以用了简单的servlet作后台业务处理,基于此,扩展了一下servlet的设计.

下面是JAVA代码:

 

package com.zwr.app.servlet;

import java.io.IOException;
import java.util.ArrayList;
import java.util.List;

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

import com.zwr.app.model.WebSiteModel;
import com.zwr.app.util.RequestParamUtil;

/**
 * extjs前台AJAX与后台交互测试Servlet
 * 
 * @author zhu wei rong
 * @since 2008-08-25
 */
@SuppressWarnings( { "serial", "unused" })
public class UserServlet extends HttpServlet {
	private final String encoding = "utf-8";

	private HttpServletRequest request;

	private HttpServletResponse response;

	protected void doBusiness() throws ServletException, IOException {
		String method = request.getParameter("method");
		if (method.equals("save")) {
			doSave();
		} else if (method.equals("list")) {
			doList();
		}
	}

	/**
	 * 测试grid分页显示数据,所有数据以json格式在JSP页面显示
	 */
	private void doList() throws ServletException, IOException {
		int start = RequestParamUtil.getIntParam(request, "start", 0);
		int limit = RequestParamUtil.getIntParam(request, "limit", 0);

		String sort = RequestParamUtil.getParam(request, "sort", null);
		String dir = RequestParamUtil.getParam(request, "dir", null);
		System.out.println(start + "\t" + limit + "\t" + sort + "\t" + dir);
		List<WebSiteModel> list = new ArrayList<WebSiteModel>();
		for (int i = 1; i <= 10; i++) {
			WebSiteModel model = new WebSiteModel();
			model.setId(i);
			model.setName("站点_" + i);
			model.setUrl("url--" + i);
			list.add(model);
		}
		int totalCount = list.size(); // 总数
		request.setAttribute("count", totalCount);
		request.setAttribute("entites", list.subList(start, (start + limit) < totalCount ? (start + limit) : totalCount));
		forward("/scripts/app/data.jsp");
	}

	private void doSave() throws ServletException, IOException {
		String name = request.getParameter("name");
		String password = request.getParameter("password");
		System.out.println("前台AJAX传递过来的参数:\t" + name + "\t" + password);
		forward("/");
	}

	@Override
	protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
		request.setCharacterEncoding(encoding);
		response.setCharacterEncoding(encoding);
		this.request = request;
		this.response = response;
		doBusiness();
	}

	@Override
	protected void doPost(HttpServletRequest arg0, HttpServletResponse arg1) throws ServletException, IOException {
		this.doGet(arg0, arg1);
	}

	public void forward(String url) throws ServletException, IOException {
		getServletContext().getRequestDispatcher(url).forward(request, response);
	}
}

 

评论

6 楼 laiseeme 2008-09-01  

那个web。xml就够你忙乎的

5 楼 stworthy 2008-09-01  

最少应该包装一下Servlet，或者做一个简单的Dispatcher。否则Servlet也太多了。

4 楼 vvgg 2008-09-01  

有多线程问题

3 楼 wangxin0072000 2008-09-01  

Struts中有个dispatch的概念，为什么不只用这种方式呢？你的这种方式耦合度太高了，很难有效的扩展。我只做过Struts，我相信几乎所有的web MVC框架都有这个功能，而且做的更有弹性和扩展性。而且实在没看出来doSave是做什么用的，而且如果目的是扩展HttpServlet，这个方法至少应该是protected的吧。

2 楼 wtusmchen 2008-09-01  

连struts都拼错了，失败

1 楼 wtusmchen 2008-09-01  

看似有很多问题：
1.一个模块一个servlet，配置起来太麻烦，建议你用spring
2.private HttpServletRequest request;在servlet里面定义变量，这个应该有多线程问题
3.Ext与后台交互是通过ajax，request.setAttribute，然后forward明显是strtus的习惯，你forward到jsp之后怎么办呢？还是要转换成josn，建议你直接在doList里面输出json，省下一个jsp文件