java - 没有名为 'springSecurityFilterChain' 的可用 bean

Question

我收到错误

No bean named 'springSecurityFilterChain' available

当我不包含 spring 安全配置时，我的代码工作正常。该项目是使用基于 spring 注解的配置创建的。当我使用 Maven pom.xml、web.xml 创建项目时，会自动创建 MvcConfiguration。现在我想使用相同的项目和配置来实现 Spring-security。我的配置文件如下所示。

MvcConfiguration.java

@Configuration
@ComponentScan(basePackages="com.mywebsite.emusicstore")
@EnableWebMvc
public class MvcConfiguration extends WebMvcConfigurerAdapter 
{

@Bean
public ViewResolver getViewResolver(){
    InternalResourceViewResolver resolver = new InternalResourceViewResolver();
    resolver.setPrefix("/WEB-INF/views/");
    resolver.setSuffix(".jsp");
    return resolver;
}

@Override
public void addResourceHandlers(ResourceHandlerRegistry registry) {
    registry.addResourceHandler("/resources/**").addResourceLocations("/resources/");
}

@Bean
public CommonsMultipartResolver multipartResolver() {
    CommonsMultipartResolver resolver=new CommonsMultipartResolver();
    Long maxUploadSize = 2048000L;
    resolver.setMaxUploadSize(maxUploadSize);
    return resolver;
}   }

SpringSecurity.java

@Configuration
@EnableWebSecurity
public class SpringSecurity extends WebSecurityConfigurerAdapter{

@Override
protected void configure(HttpSecurity http) throws Exception {

      http.authorizeRequests()
        .antMatchers("/").permitAll()
        .antMatchers("/admin**").access("ROLE_USER")
        .and().formLogin().loginPage("/login").defaultSuccessUrl("/admin/").failureUrl("/login?error").usernameParameter("username").passwordParameter("password")
        .and().logout().logoutSuccessUrl("/login?logout");  

      //http.csrf().disable();
    }  

    @Autowired
    public void configureGlobal(AuthenticationManagerBuilder auth) throws Exception {
        HibernateConfig hc = new HibernateConfig();
        String authoritiesByUsernameQuery = "SELECT username,authorities FROM authorities WHERE username = ?";
        String usersByUsernameQuery = "SELECT username,password,enabled FROM users WHERE username = ?";
        auth.jdbcAuthentication().dataSource(hc.dataSource()).authoritiesByUsernameQuery(authoritiesByUsernameQuery).usersByUsernameQuery(usersByUsernameQuery);
    }

}

web.xml

<?xml version="1.0" encoding="UTF-8"?>
<web-app version="2.4" xmlns="http://java.sun.com/xml/ns/j2ee"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://java.sun.com/xml/ns/j2ee 
http://java.sun.com/xml/ns/j2ee/web-app_2_4.xsd">
<display-name>emusicstore</display-name>
<context-param>
<param-name>contextClass</param-name>
<param-value>
        org.springframework.web.context.support.AnnotationConfigWebApplicationContext
    </param-value>
</context-param>
<listener>
    <listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
</listener>

<servlet>
    <servlet-name>SpringDispatcher</servlet-name>
    <servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
    <init-param>
        <param-name>contextClass</param-name>
        <param-value>
            org.springframework.web.context.support.AnnotationConfigWebApplicationContext
        </param-value>
    </init-param>
    <init-param>
        <param-name>contextConfigLocation</param-name>
        <param-value>com.mywebsite.emusicstore</param-value>
    </init-param>
    <load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
    <servlet-name>SpringDispatcher</servlet-name>
    <url-pattern>/</url-pattern>
</servlet-mapping>
<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>

<session-config>
    <session-timeout>30</session-timeout>
</session-config>

错误

SEVERE: Exception starting filter springSecurityFilterChain
org.springframework.beans.factory.NoSuchBeanDefinitionException: No bean 
named 'springSecurityFilterChain' available
at org.springframework.beans.factory.support.DefaultListableBeanFactory.getBeanDefinition(DefaultListableBeanFactory.java:687)
at org.springframework.beans.factory.support.AbstractBeanFactory.getMergedLocalBeanDefinition(AbstractBeanFactory.java:1207)
at org.springframework.beans.factory.support.AbstractBeanFactory.doGetBean(AbstractBeanFactory.java:284)
at org.springframework.beans.factory.support.AbstractBeanFactory.getBean(AbstractBeanFactory.java:202)
at org.springframework.context.support.AbstractApplicationContext.getBean(AbstractApplicationContext.java:1086)
at org.springframework.web.filter.DelegatingFilterProxy.initDelegate(DelegatingFilterProxy.java:327)
at org.springframework.web.filter.DelegatingFilterProxy.initFilterBean(DelegatingFilterProxy.java:235)
at org.springframework.web.filter.GenericFilterBean.init(GenericFilterBean.java:236)
at org.apache.catalina.core.ApplicationFilterConfig.initFilter(ApplicationFilterConfig.java:279)
at org.apache.catalina.core.ApplicationFilterConfig.getFilter(ApplicationFilterConfig.java:260)
at org.apache.catalina.core.ApplicationFilterConfig.<init>(ApplicationFilterConfig.java:105)
at org.apache.catalina.core.StandardContext.filterStart(StandardContext.java:4598)
at org.apache.catalina.core.StandardContext.startInternal(StandardContext.java:5223)
at org.apache.catalina.util.LifecycleBase.start(LifecycleBase.java:155)
at org.apache.catalina.core.ContainerBase$StartChild.call(ContainerBase.java:1404)
at org.apache.catalina.core.ContainerBase$StartChild.call(ContainerBase.java:1394)
at java.util.concurrent.FutureTask.run(Unknown Source)
at java.util.concurrent.ThreadPoolExecutor.runWorker(Unknown Source)
at java.util.concurrent.ThreadPoolExecutor$Worker.run(Unknown Source)
at java.lang.Thread.run(Unknown Source)

Answer 1

请也添加此类。

import org.springframework.security.web.context.AbstractSecurityWebApplicationInitializer;

public class SecurityWebApplicationInitializer extends AbstractSecurityWebApplicationInitializer{

}