Question

我正在将Spring4与tomcat一起使用，我的web.xml如下：

<context-param>
    <param-name>contextConfigLocation</param-name>
    <param-value>WEB-INF/classes/resources/beans.xml</param-value>
</context-param>
<listener>
    <listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
</listener>

<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>

<servlet>
    <servlet-name>spring</servlet-name>
    <servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
    <init-param>
        <param-name>contextConfigLocation</param-name>
        <param-value>/WEB-INF/spring-servlet.xml</param-value>
    </init-param>
    <load-on-startup>1</load-on-startup>
</servlet>

<servlet-mapping>
    <servlet-name>spring</servlet-name>
    <url-pattern>*.html</url-pattern>
</servlet-mapping>
<resource-ref>
    <description>MySQL Datasource</description>
    <res-ref-name>jdbc/informagiovani</res-ref-name>
    <res-type>javax.sql.DataSource</res-type>
    <res-auth>Container</res-auth>
</resource-ref>
<welcome-file-list>
    <welcome-file>index.html</welcome-file>
</welcome-file-list>

我的beans.xml如下所示，并且为空，因为我需要的所有bean都通过注释定义：

<beans xmlns = "http://www.springframework.org/schema/beans"
xmlns:xsi = "http://www.w3.org/2001/XMLSchema-instance"
xmlns:context = "http://www.springframework.org/schema/context"
xsi:schemaLocation = "http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans-3.0.xsd
    http://www.springframework.org/schema/context http://www.springframework.org/schema/context/spring-context.xsd">

<context:annotation-config />

</beans>

我定义 spring-servlet.xml 如下：

<?xml version="1.0" encoding="UTF-8"?>
<beans xmlns="http://www.springframework.org/schema/beans"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xmlns:p="http://www.springframework.org/schema/p"
xmlns:context="http://www.springframework.org/schema/context"
xmlns:mvc="http://www.springframework.org/schema/mvc"
xsi:schemaLocation="http://www.springframework.org/schema/beans
http://www.springframework.org/schema/beans/spring-beans-3.0.xsd
http://www.springframework.org/schema/context
http://www.springframework.org/schema/context/spring-context-3.0.xsd
http://www.springframework.org/schema/mvc http://www.springframework.org/schema/mvc/spring-mvc-3.0.xsd">

<context:component-scan base-package="it.informagiovani" />

<mvc:annotation-driven />

<bean id="viewResolver" class="org.springframework.web.servlet.view.UrlBasedViewResolver">
    <property name="viewClass" value="org.springframework.web.servlet.view.JstlView" />
    <property name="prefix" value="/WEB-INF/views/" />
    <property name="suffix" value=".jsp" />
</bean>

</beans>

我在Spring文档中找到了安全配置：

@Configuration
@EnableWebSecurity
public class SecurityConfiguration extends WebSecurityConfigurerAdapter {

@Autowired
public void configureGlobalSecurity(AuthenticationManagerBuilder auth) throws Exception {
    auth.inMemoryAuthentication().withUser("giacomo").password("230483").roles("ADMIN");
}

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

  http.authorizeRequests()
    .antMatchers("/", "/home").permitAll() 
    .antMatchers("/addUser").access("hasRole('ADMIN')")
    .antMatchers("/saveUser").access("hasRole('ADMIN')")
    .and().formLogin();


}
}

启动复制时，我得到：

No bean named 'springSecurityFilterChain' available

怎么了？我想我至少在弹簧安全性方面混合了注释和xml配置，但是由于先前的配置，我需要web.xml文件。有没有办法解决这个问题，或者我必须使用 security-config.xml 而不是我发布的配置类？

谢谢

Answer 1

我认为问题出在您的beans.xml配置文件中。您没有声明component-scan属性，并且Spring不知道您具有Java安全配置。

只需将<context:component-scan base-package="my.package"/>添加到beans.xml，错误就会消失。

Spring4安全配置

1 个答案: