首先看看web.xml里的写法吧
xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_3_0.xsd">
要用到的包就是struts1里的包,我这里用的是这几个
Antlr.jar
Commons-beanutils.jar
Commons-digester.jar
Commons-fileupload.jar
Commons-logging.jar
Commons-validator.jar
Jakarta-oro.jar
Struts.jar
实体类要继承ActionForm这个类:
import org.apache.struts.action.ActionForm;
public class LoginForm extends ActionForm {
private String username;
private String password;
public String getUsername() {
return username;
}
public void setUsername(String username) {
this.username = username;
}
public String getPassword() {
return password;
}
public void setPassword(String password) {
this.password = password;
}
}
控制层里的类要继承Action这个类,而且要重写execute这个方法
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;
import org.apache.struts.action.Action;
import org.apache.struts.action.ActionForm;
import org.apache.struts.action.ActionForward;
import org.apache.struts.action.ActionMapping;
public class LoginAction extends Action {
@Override
public ActionForward execute(ActionMapping mapping, ActionForm form,
HttpServletRequest request, HttpServletResponse response)
throws Exception {
//需要将form转型成真正的类型,这样才可以取得浏览器的参数信息
LoginForm lf = (LoginForm) form;
if(lf.getUsername().equals(lf.getPassword())){//参数和在 return mapping.findForward("loginSuccess"); }else{ return mapping.findForward("loginFailure"); } } } 首页里的写法: <%@ page language="java" import="java.util.*" pageEncoding="UTF-8"%> <% String path = request.getContextPath(); String basePath = request.getScheme() + "://" + request.getServerName() + ":" + request.getServerPort() + path + "/"; %> struts1中的struts-config.xml中的配置 最后就是把这个项目部署到服务器里运行了,在action层里我加的判断是输入的用户名跟密码一样的话就登录成功,不一样的话就登录失败 祝你好运。下载本文