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Home/ Questions/Q 4126084
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Editorial Team
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Editorial Team
Asked: May 20, 20262026-05-20T23:57:15+00:00 2026-05-20T23:57:15+00:00

Error Context initialization failed with Spring3 Security using JDBC. File: applicationContext-security-JDBC.xml <?xml version=1.0 encoding=UTF-8?>

  • 0

Error Context initialization failed with Spring3 Security using JDBC.

File:applicationContext-security-JDBC.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"
    xmlns:jdbc="http://www.springframework.org/schema/jdbc"
    xsi:schemaLocation="http://www.springframework.org/schema/beans
           http://www.springframework.org/schema/beans/spring-beans-3.0.xsd
           http://www.springframework.org/schema/jdbc
           http://www.springframework.org/schema/jdbc/spring-jdbc-3.0.xsd
           http://www.springframework.org/schema/security
           http://www.springframework.org/schema/security/spring-security-3.0.xsd">


    <http auto-config="true">
        <intercept-url pattern="/friends/**" access="ROLE_USER" />
        <intercept-url pattern="/articles/**" access="ROLE_USER" />
        <intercept-url pattern="/**" access="IS_AUTHENTICATED_ANONYMOUSLY" />
    </http>

    <authentication-manager alias="authenticationManager">

        <authentication-provider>
            <jdbc-user-service data-source-ref="dataSource" />
        </authentication-provider>
    </authentication-manager>
</beans:beans>

Web.XML

<?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">

<listener>
       <listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
    </listener>

    <context-param>
         <param-name>contextConfigLocation</param-name>
         <param-value>
        /WEB-INF/applicationContext-security-JDBC.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>


    <context-param>
        <param-name>log4jConfigLocation</param-name>
        <param-value>/WEB-INF/classes/log4j.properties</param-value>
    </context-param>



    <servlet>
        <servlet-name>dispatcher</servlet-name>
        <servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
        <load-on-startup>1</load-on-startup>
    </servlet>

    <servlet-mapping>
        <servlet-name>dispatcher</servlet-name>
        <url-pattern>*.html</url-pattern>
    </servlet-mapping>

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



    <!--
      - Loads the root application context of this web app at startup.
      - The application context is then available via
      - WebApplicationContextUtils.getWebApplicationContext(servletContext).
    -->
    <listener>
        <listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
    </listener>

    <!--
      - Publishes events for session creation and destruction through the application
      - context. Optional unless concurrent session control is being used.
      -->
    <listener>
      <listener-class>org.springframework.security.web.session.HttpSessionEventPublisher</listener-class>
    </listener>

    <listener>
        <listener-class>org.springframework.web.util.Log4jConfigListener</listener-class>
    </listener>


</web-app>

Below is the error:

[ERROR,ContextLoader] Context initialization failed
org.springframework.beans.factory.BeanCreationException: Error creating bean with name ‘org.springframework.security.authentication.ProviderManager#0’: Cannot create inner bean ‘(inner bean)’ of type [org.springframework.security.config.authentication.AuthenticationManagerFactoryBean] while setting bean property ‘parent’; nested exception is org.springframework.beans.factory.BeanCreationException: Error creating bean with name ‘(inner bean)’: FactoryBean threw exception on object creation; nested exception is org.springframework.beans.factory.BeanCreationException: Error creating bean with name ‘org.springframework.security.authenticationManager’: Cannot resolve reference to bean ‘org.springframework.security.authentication.dao.DaoAuthenticationProvider#0’ while setting bean property ‘providers’ with key [0]; nested exception is org.springframework.beans.factory.BeanCreationException: Error creating bean with name ‘org.springframework.security.authentication.dao.DaoAuthenticationProvider#0’: Cannot resolve reference to bean ‘org.springframework.security.provisioning.JdbcUserDetailsManager#0’ while setting bean property ‘userDetailsService’; nested exception is org.springframework.beans.factory.BeanCreationException: Error creating bean with name ‘org.springframework.security.provisioning.JdbcUserDetailsManager#0’: Cannot resolve reference to bean ‘dataSource’ while setting bean property ‘dataSource’; nested exception is org.springframework.beans.factory.NoSuchBeanDefinitionException: No bean named ‘dataSource’ is defined

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  1. Editorial Team
    Editorial Team
    2026-05-20T23:57:17+00:00Added an answer on May 20, 2026 at 11:57 pm

    That is easy to answer. The exception message says (at the end) “No bean named ‘dataSource’ is defined”. And sure enough, your applicationContext-security-JDBC.xml file does not include a definition for a dataSource bean.

    Section 6.2 of the SpringSecurity manual includes material on how to configure a data source for a JDBC user details service. (Hint: SpringSecurity needs to be told what database, what username and what password to use …)

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