spring4+hibernate4+struts2注解，class找不到bean

程序员文章站 2022-06-05 14:17:07

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最近想整合S2SH，采用的是spring4+hibernate4+struts2+maven+MySQL+jdk1.8+tomcat7.

整个项目采用注解实现：spring、hibernate都采用注解实现.

struts.xml:

<constant name="struts.objectFactory" value="spring"/>
<constant name="struts.devMode" value="true"/>
<constant name="struts.configuration.xml.reload" value="true"/>
<constant name="struts.serve.static.browserCache" value="false"/>
<constant name="struts.i18n.encoding" value="UTF-8"/>
<constant name="struts.i18n.reload" value="true"/>
<constant name="struts.multipart.maxSize" value="104857600"/>
<constant name="struts.enable.DynamicMethodInvocation" value="false"/>
 
<package name="common" extends="struts-default"></package>
 
<package name="login" namespace="/login"     extends="common">
    <action name="login" class="userLogin">
        <result name="loginFail">fail.jsp</result>
        <result name="success">index.jsp</result>
    </action>
</package>

spring.xml:

<context:annotation-config/>
	<context:component-scan base-package="com.value.yun.modules" />

controller中的java代码：

package com.value.yun.modules.controller;

import com.opensymphony.xwork2.ActionSupport;
import com.value.yun.common.base.Encryption;
import com.value.yun.modules.entity.User;
import com.value.yun.modules.service.UserService;
import com.value.yun.utils.StringUtils;
import org.apache.struts2.ServletActionContext;
import org.springframework.beans.factory.annotation.Autowired;
import org.springframework.stereotype.Controller;
import org.springframework.web.bind.annotation.RequestMapping;

import javax.servlet.http.HttpServletRequest;

/**
 * Created by 丽 on 2015/4/16.
 */
@Controller
public class LoginController extends ActionSupport{
    public LoginController(){
        System.out.println("============login controller constructor==========================");
    }
    private User user;
    @Autowired
    private UserService userService;

    public User getUser() {
        return user;
    }

    public void setUser(User user) {
        this.user = user;
    }

    @RequestMapping(value = "userLogin")
    public String login(){
        HttpServletRequest request = ServletActionContext.getRequest();
        if (user==null){
            request.setAttribute("loginFail","对象为空");
            return "loginFail";
        }
        if(StringUtils.isBlank(user.getLoginName())){
            request.setAttribute("loginFail","登录账号为空");
            return "loginFail";
        }
        if (StringUtils.isBlank(user.getPassword())){
            request.setAttribute("loginFail","登录密码为空");
            return "loginFail";
        }
        request.setAttribute("loginName",user.getLoginName());
        request.setAttribute("password", Encryption.encrytMD532(user.getPassword()));
        return "success";
    }
}

上面无参构造方法是有执行的，在启动tomcat的时候执行的。

index.jsp:

<%@ page language="java" contentType="text/html; charset=UTF-8" pageEncoding="UTF-8"%>
<html>
<head>
    <meta name="decorator" content="default"/>
    <title>欢迎来到登录界面</title>
</head>
<body>
    <h2>Hello World!</h2>
    <form action="${baseUrl}/login/login" method="post" class="loginForm">
        <table>
            <tr>
                <th>账号：</th>
                <td><input type="text" name="user.loginName"></td>
            </tr>
            <tr>
                <th>密码：</th>
                <td><input type="password" name="user.password"></td>
            </tr>
            <tr>
                <td><button type="submit">登录</button></td>
                <td><button type="reset">重置</button></td>
            </tr>
        </table>
    </form>

${loginName}<br>
${password}
</body>
</html>

在点击登陆的时候，他就提示<action name="login" class="userLogin">这一行出错。

具体信息如下：

Unable to instantiate Action, userLogin, defined for 'login' in namespace '/login'userLogin - action - file:/D:/yun_systems/hua_yu/source/EASMS/target/EASMS/WEB-INF/classes/struts/login.xml:8:48

请问，有没有人知道这是怎么回事啊？

相关标签： Unable to instantiate Action

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