在Spring中添加管理部分

时间：2013-03-08 20:26:35

标签： java spring spring-mvc

我有一个Spring网络项目，需要一个管理部分。我认为这部分很容易，我会遇到安全问题，但我甚至无法指向/ admin /部分。

我在dispatcher-servelet.xml中有以下内容将JSP文件映射到控制器：

<context:component-scan base-package="controller"/>
<context:component-scan base-package="controller.admin"/>

<bean id="viewResolver"
      class="org.springframework.web.servlet.view.InternalResourceViewResolver"
      p:prefix="/WEB-INF/jsp/"
      p:suffix=".jsp" />

现在/ WEB-INF / jsp /中有一个标有“admin”的文件夹，我有adminindex.jsp。我在web.xml中也有以下内容

<?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">
    <context-param>
        <param-name>contextConfigLocation</param-name>
        <param-value>/WEB-INF/applicationContext.xml</param-value>
    </context-param>
    <listener>
        <listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
    </listener>
    <servlet>
        <servlet-name>dispatcher</servlet-name>
        <servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
        <load-on-startup>2</load-on-startup>
    </servlet>
    <servlet-mapping>
        <servlet-name>dispatcher</servlet-name>
        <url-pattern>*.htm</url-pattern>
    </servlet-mapping>
    <session-config>
        <session-timeout>
            30
        </session-timeout>
    </session-config>
    <welcome-file-list>
        <welcome-file>redirect.jsp</welcome-file>
    </welcome-file-list>
    <security-constraint>
        <display-name>Constraint1</display-name>
    </security-constraint>
    <security-constraint>
        <display-name>admin pages</display-name>
        <web-resource-collection>
            <web-resource-name>Administration Pages</web-resource-name>
            <description/>
            <url-pattern>/admin/*</url-pattern>
        </web-resource-collection>
        <auth-constraint>
            <description/>
            <role-name>eCommerceAdmin</role-name>
        </auth-constraint>
        <!--  <user-data-constraint>
            <description/>
            <transport-guarantee>CONFIDENTIAL</transport-guarantee>
        </user-data-constraint>-->
    </security-constraint>
</web-app>

当我尝试访问

时

本地主机：8080 / NewWebsite /管理/ adminindex.htm

，我得到了

  

拒绝访问所请求的资源

     

INFO：ContextListener：   attributeAdded（ 'org.apache.jasper.compiler.TldLocationsCache'，   'org.apache.jasper.compiler.TldLocationsCache@44d1bd08'）

我可以轻松访问

localhost：8080 / NewWebsite / index.htm，我还想为admin文件夹设置自动重定向，就像使用根文件夹一样。 I.E去

localhost：8080 / NewWebsite /指向index.htm。

任何帮助都会很棒。

4 个答案:

答案 0 :(得分：2)

首先你需要验证用户吗？不是吗？否则，您的应用程序将如何识别ADMIN尝试访问或正常USER？
在您执行此操作之前，请从 web.xml 中删除security-constraint 所以在你的应用程序中添加spring认证 首先创建一个pojo类，以获得应该实现GrantedAuthority的{​​{1}}列表。以下是一个示例：

org.springframework.security.core.userdetails.UserDetails

以下是您需要的HTTP标记。

public class YourPojo implements UserDetails{


    /** The authorities. */
    //This collection will have eCommerceAdmin
    public Collection<GrantedAuthority> authorities;

    /** The username. */
    public String username;

    /** The account non expired. */
    public boolean accountNonExpired;

    /** The credentials non expired. */
    public boolean credentialsNonExpired;

    /** The enabled. */
    public boolean enabled;

    /** The Constant serialVersionUID. */
    private static final long serialVersionUID = -2342376103893073629L;

    /* (non-Javadoc)
     * @see org.springframework.security.core.userdetails.UserDetails#getAuthorities()
     */
    @Override
    public Collection<GrantedAuthority> getAuthorities() {
        return authorities;
    }

    /* (non-Javadoc)
     * @see org.springframework.security.core.userdetails.UserDetails#getPassword()
     */
    @Override
    public String getPassword() {
        return null;
    }

    /* (non-Javadoc)
     * @see org.springframework.security.core.userdetails.UserDetails#getUsername()
     */
    @Override
    public String getUsername() {
        return username;
    }

    /* (non-Javadoc)
     * @see org.springframework.security.core.userdetails.UserDetails#isAccountNonExpired()
     */
    @Override
    public boolean isAccountNonExpired() {
        return accountNonExpired;
    }

    /* (non-Javadoc)
     * @see org.springframework.security.core.userdetails.UserDetails#isAccountNonLocked()
     */
    @Override
    public boolean isAccountNonLocked() {
        return accountNonLocked;
    }

    /* (non-Javadoc)
     * @see org.springframework.security.core.userdetails.UserDetails#isCredentialsNonExpired()
     */
    @Override
    public boolean isCredentialsNonExpired() {
        return credentialsNonExpired;
    }

    /* (non-Javadoc)
     * @see org.springframework.security.core.userdetails.UserDetails#isEnabled()
     */
    @Override
    public boolean isEnabled() {
        return enabled;
    }


    /**
     * Sets the authorities.
     *
     * @param authorities the new authorities
     */
    public void setAuthorities(Collection<GrantedAuthority> authorities) {
        this.authorities = authorities;
    }

    /**
     * Sets the username.
     *
     * @param username the new username
     */
    public void setUsername(String username) {
        this.username = username;
    }

    /**
     * Sets the account non expired.
     *
     * @param accountNonExpired the new account non expired
     */
    public void setAccountNonExpired(boolean accountNonExpired) {
        this.accountNonExpired = accountNonExpired;
    }

    /**
     * Sets the account non locked.
     *
     * @param accountNonLocked the new account non locked
     */
    public void setAccountNonLocked(boolean accountNonLocked) {
        this.accountNonLocked = accountNonLocked;
    }

    /**
     * Sets the credentials non expired.
     *
     * @param credentialsNonExpired the new credentials non expired
     */
    public void setCredentialsNonExpired(boolean credentialsNonExpired) {
        this.credentialsNonExpired = credentialsNonExpired;
    }

    /**
     * Sets the enabled.
     *
     * @param enabled the new enabled
     */
    public void setEnabled(boolean enabled) {
        this.enabled = enabled;
    }

}

现在定义您的身份验证提供程序。

<!-- to use Spring security tags -->
    <bean class="org.springframework.security.web.access.expression.DefaultWebSecurityExpressionHandler" />
<http pattern="/login*" security="none"/>
<http pattern="/static/**" security="none"/>        
<http auto-config="false">      
        <intercept-url pattern="/admin/**" access="eCommerceAdmin" />
        <form-login login-page="/login" default-target-url="/welcome"
            authentication-failure-url="/loginfailed" />
        <logout logout-success-url="/logout" />     

        <session-management>
            <concurrency-control max-sessions="1" error-if-maximum-exceeded="true" />
        </session-management>           
    </http>

此<bean id="customeAuthProvider" class="your.auth.provider.class"> </bean> <authentication-manager > <authentication-provider ref="customeAuthProvider" ></authentication-provider> </authentication-manager> 应实施customeAuthProvider。

org.springframework.security.authentication.AuthenticationProvider

FYR您可以在web.xml中引用security xml文件，如下所示..而且您的web.xml也应该有spring安全过滤器。

@Override
    public Authentication authenticate(Authentication authentication)
            throws AuthenticationException {

        UsernamePasswordAuthenticationToken userToken = (UsernamePasswordAuthenticationToken)authentication;
        String username = userToken.getName();
        String password = (String) authentication.getCredentials();
          //Do whatevr you want with the credentials
         //Then populate the authorities for this credential
         YourPojo user=new YourPojo ();
         user.setUserName("add username");
        //set other details
        List<GrantedAuthority> grantedAuthorityList = new ArrayList<GrantedAuthority>();
        //if user is admin add the below line
        GrantedAuthorityImpl grantedAuthorityImpl = new GrantedAuthorityImpl("eCommerceAdmin");
       //Add other authorities as applicable like 'user' etc.
       user.setAuthorities(grantedAuthorityList);
       return new UsernamePasswordAuthenticationToken(username, password, user.getAuthorities());

您还需要Spring安全依赖项。如果您在项目中使用Maven，请添加以下依赖项，否则您可以手动下载这些jar并继续。

<context-param>
 <param-name>contextConfigLocation</param-name>
  <param-value>
    /WEB-INF/your-applicationContext.xml
    /WEB-INF/your-spring-security.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>

现在你很高兴.. FYR通过this

答案 1 :(得分：0)

您是否尝试将前缀更改为WEB-INF / jsp / admin /？

答案 2 :(得分：0)

问题不在viewresolver中，因此您提供的信息不足以回答您的问题。请显示更多配置，尤其是安全配置。

除了你的问题。 controller.admin上的组件扫描不一定是必要的，因为第一个扫描它。

答案 3 :(得分：0)

我认为你应该考虑使用spring security，然后你可以配置spring security role系统来控制访问。下面是从我的一个应用程序中获取的片段，因为您可以看到/ admin路径要求调用者具有Admin角色。 Spring安全性设置有点复杂，但一旦设置它就能很好地工作。

<http auto-config='true' use-expressions="true" >

    <!-- public resources that everyone should be able to access -->
    <intercept-url pattern="/favicon.ico" access="permitAll" />
    <intercept-url pattern="/login"       access="permitAll" />
    <intercept-url pattern="/login/error" access="permitAll" />
    <intercept-url pattern="/**/*.js"        access="permitAll" />
    <intercept-url pattern="/**/*.jsp"       access="denyAll" />

    <intercept-url pattern="/admin/**"     access="hasRole('admin')" />

    </session-management>

</http>
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