java.lang.IllegalStateException: BeanFactory 未初始化或已关闭 - 在通过 ApplicationContext 访问 bean 之前调用“刷新”
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java.lang.IllegalStateException: BeanFactory not initialized or already closed - call 'refresh' before accessing beans via the ApplicationContext
提问by Hyman
I need to add Spring Security with customized login page and connection to database to my Spring MVC project. I am receiving following error message, based on answers of other questions, I tried to change the code, for example I changed my Spring Security Schema version to 4.0 but the code returns following error:
我需要将带有自定义登录页面和数据库连接的 Spring Security 添加到我的 Spring MVC 项目。我收到以下错误消息，根据其他问题的答案，我尝试更改代码，例如我将 Spring Security Schema 版本更改为 4.0，但代码返回以下错误：
Changed schema to 4.0
将架构更改为 4.0
http://www.springframework.org/schema/security/spring-security-4.0.xsd
Error
错误
Cannot initialize context because there is already a root application context
present - check whether you have multiple ContextLoader* definitions in your
web.xml!
My code
我的代码
my-security.xml
我的安全.xml
<?xml version="1.0" encoding="UTF-8"?>
<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-3.2.xsd
http://www.springframework.org/schema/security
http://www.springframework.org/schema/security/spring-security-3.2.xsd'>
<beans:import resource="security-db.xml" />
<http auto-config="true" access-denied-page="/notFound.jsp"
use-expressions="true">
<intercept-url pattern="/" access="permitAll" />
</http>
</beans:beans>
web.xml
网页.xml
<?xml version="1.0" encoding="UTF-8"?>
<web-app 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_3_0.xsd"
version="3.0">
<listener>
<listener-class>org.apache.tiles.extras.complete.CompleteAutoloadTilesListener</listener-class>
</listener>
<listener>
<listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
</listener>
<servlet>
<servlet-name>my</servlet-name>
<servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>my</servlet-name>
<url-pattern>/</url-pattern>
</servlet-mapping>
<context-param>
<param-name>contextConfigLocation</param-name>
<param-value>/WEB-INF/config/my-security.xml
</param-value>
</context-param>
</web-app>
security-db.xml
安全-db.xml
<beans xmlns="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-3.1.xsd">
<bean id="dataSource"
class="org.springframework.jdbc.datasource.DriverManagerDataSource">
<property name="driverClassName" value="com.mysql.jdbc.Driver" />
<property name="url" value="jdbc:mysql://localhost/dbproj" />
<property name="username" value="Hyman" />
<property name="password" value="Hyman" />
</bean>
</beans>
采纳答案by Sheetal Mohan Sharma
I think you need have only one xml configuration file (my-servlet.xml as your servlet name is "my"so filename must be "my-servlet.xml") in web-xml and then refer others in that file. Refer to xmls below.
我认为您只需要在 web-xml 中有一个 xml 配置文件（my-servlet.xml，因为您的servlet 名称是“my”，所以文件名必须是“ my-servlet.xml”），然后在该文件中引用其他文件。参考下面的xmls。
<xml version="1.0" encoding="UTF-8"?>
<web-app 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_3_0.xsd"
version="3.0">
<listener>
<listener-class>org.apache.tiles.extras.complete.CompleteAutoloadTilesListener</listener-class>
</listener>
<listener>
<listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
</listener>
<servlet>
<servlet-name>my</servlet-name>
<servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
<init-param>
<param-name>contextConfigLocation</param-name>
<param-value>/WEB-INF/my-servlet.xml</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
</web-app>
In my-servlet.xml file you can use import to compose your other XML configurations.
在 my-servlet.xml 文件中，您可以使用 import 来编写其他 XML 配置。
<beans>
<bean id="bean1" class="..."/>
<bean id="bean2" class="..."/>
<import resource="security-db.xml"/>
<import resource="foo-db.xml"/>
</beans>
回答by Ali Dehghani
Add DelegatingFilterProxyto your web.xml:
添加DelegatingFilterProxy到您的 web.xml：
<?xml version="1.0" encoding="UTF-8"?>
<web-app xmlns="http://xmlns.jcp.org/xml/ns/javaee"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://xmlns.jcp.org/xml/ns/javaee
http://xmlns.jcp.org/xml/ns/javaee/web-app_3_1.xsd"
version="3.1">
<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>
<listener>
<listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
</listener>
<servlet>
<servlet-name>my</servlet-name>
<servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>my</servlet-name>
<url-pattern>/</url-pattern>
</servlet-mapping>
<context-param>
<param-name>contextConfigLocation</param-name>
<param-value>classpath:config/security-config.xml</param-value>
</context-param>
</web-app>
add my-servlet.xmlas your web application context configuration to /webapp/WEB-INF/.
add authentication-managerelement to my-security.xml:
添加my-servlet.xml作为您的 Web 应用程序上下文配置到/webapp/WEB-INF/. 添加authentication-manager元素到my-security.xml：
<?xml version="1.0" encoding="UTF-8"?>
<beans:beans xmlns:beans="http://www.springframework.org/schema/beans"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xmlns="http://www.springframework.org/schema/security"
xsi:schemaLocation="http://www.springframework.org/schema/beans
http://www.springframework.org/schema/beans/spring-beans.xsd
http://www.springframework.org/schema/security
http://www.springframework.org/schema/security/spring-security.xsd">
<beans:import resource="spring-db.xml" />
<http auto-config="true" use-expressions="true">
<intercept-url pattern="/" access="permitAll" />
</http>
<authentication-manager>
<authentication-provider>
<user-service>
<user name="username" authorities="ROLE_ADMIN" password="password" />
</user-service>
</authentication-provider>
</authentication-manager>
</beans:beans>
get rid of access-denied-pageand use access-denied-handleras Mkyong did it here
摆脱access-denied-page和使用access-denied-handler作为Mkyong在这里做到了
