使用CustomUserDetailsS​​ervice进行春季安全性登录

Question

嗨，我正在尝试为用户登录实施Spring Security。 我的spring-security.xml文件是

<beans:beans xmlns="http://www.springframework.org/schema/security"
    xmlns:beans="http://www.springframework.org/schema/beans" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
    xsi:schemaLocation="http://www.springframework.org/schema/beans

http://www.springframework.org/schema/beans/spring-beans-3.0.xsd


http://www.springframework.org/schema/security


http://www.springframework.org/schema/security/spring-security-3.1.xsd">

    <http use-expressions="true">
        <intercept-url pattern="/home/**" access="isAuthenticated()" />
        <intercept-url pattern="/login" access="permitAll" />
        <intercept-url pattern="/" access="permitAll" />
        <form-login  login-page="/cart"  authentication-failure-url="/#/login?error=1" always-use-default-target="true"/>
        <logout />
    </http>

    <authentication-manager alias="authenticationManager">
        <authentication-provider user-service-ref="userDetailsService"/>
    </authentication-manager>

    <beans:bean id="userDetailsService" class="com.dashboard.service.CustomUserDetailsService"/>

</beans:beans>

这是我的CustomUserDetailsS​​ervice类

package com.dashboard.service;

import java.util.Collections;

import javax.annotation.Resource;

import org.springframework.beans.factory.annotation.Autowired;
import org.springframework.security.authentication.UsernamePasswordAuthenticationToken;
import org.springframework.security.core.authority.SimpleGrantedAuthority;
import org.springframework.security.core.userdetails.User;
import org.springframework.security.core.userdetails.UserDetails;
import org.springframework.security.core.userdetails.UserDetailsService;
import org.springframework.security.core.userdetails.UsernameNotFoundException;
import org.springframework.stereotype.Service;
import static java.util.Arrays.asList;

import com.dashboard.repositories.UserRepository;

@Service
public class CustomUserDetailsService implements UserDetailsService {

    @Resource
    UserService userService;

    @Autowired
    UserRepository userRepository;

    @Override
    public UserDetails loadUserByUsername(String email)     throws UsernameNotFoundException {
        System.out.println("here");
            try{
                SimpleGrantedAuthority simpleGrantedAuthority = new SimpleGrantedAuthority("ROLE_ADMIN");
                com.dashboard.Model.User userObj = userService.getUser(email);
                User user = new User(userObj.getUserName(), userObj.getPassword(), true, true, true, true, asList(simpleGrantedAuthority));
            return null;

        }catch(Exception e){
            e.printStackTrace();
            return null;
        }
        //return null;
    }
}

这是我的UserService类别

package com.dashboard.service;

import javax.annotation.Resource;

import org.springframework.stereotype.Service;

import com.dashboard.Model.User;
import com.dashboard.repositories.UserRepository;

@Service
public class UserService {

    @Resource
    UserRepository userRepository;

    public User saveUser(User user){

        return userRepository.save(user);
    }

    public User getUser(String email){
        return userRepository.findByEmail(email);
    }

}

这是我的web.xml文件

<?xml version="1.0" encoding="UTF-8"?>
<web-app version="2.5" xmlns="http://java.sun.com/xml/ns/javaee"
    xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
    xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd">

    <!-- The definition of the Root Spring Container shared by all Servlets and Filters -->
    <context-param>
        <param-name>contextConfigLocation</param-name>
        <param-value>/WEB-INF/spring/root-context.xml</param-value>
    </context-param>

 <filter>
  <filter-name>springSecurityFilterChain</filter-name>
  <filter-class>org.springframework.web.filter.DelegatingFilterProxy</filter-class>
 </filter>

 <filter-mapping>
  <filter-name>springSecurityFilterChain</filter-name>
  <url-pattern>/*</url-pattern>
 </filter-mapping>

    <!-- Creates the Spring Container shared by all Servlets and Filters -->
    <listener>
        <listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
    </listener>

    <!-- Processes application requests -->
    <servlet>
        <servlet-name>appServlet</servlet-name>
        <servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
        <init-param>
            <param-name>contextConfigLocation</param-name>
            <param-value>
            /WEB-INF/spring/appServlet/servlet-context.xml, 
            /WEB-INF/spring/spring-security.xml
            </param-value>
        </init-param>
        <load-on-startup>1</load-on-startup>
    </servlet>

    <servlet-mapping>
        <servlet-name>appServlet</servlet-name>
        <url-pattern>/</url-pattern>
    </servlet-mapping>

    <welcome-file-list>
    <welcome-file>index.html</welcome-file>
    </welcome-file-list>

</web-app>

我的问题是当我提交表格并调试应用程序时

 com.dashboard.Model.User userObj = userService.getUser(email);

我的'userService'对象为null，任何人都可以告诉我为什么bean为null。 据我了解，我已经在UserService类的开头声明了一个@service注释，那么它不应该创建一个新的bean对象吗？

Answer 1

只有声明了<context:component-scan>或<context:annotation-config>时，才会处理@Autowired注释的字段。 您将需要添加其中之一。

---

还要注意，用@Service注释类并对其进行component-scan将为该类创建一个bean。

@Service
public class CustomUserDetailsService implements UserDetailsService {

为该类放置一个bean定义

<beans:bean id="userDetailsService" class="com.dashboard.service.CustomUserDetailsService"/>

还会生成一个bean。 在这种情况下，您的上下文中将有2个类型为CustomUserDetailsService bean。 您可能不想要这个。