Question

我正在使用Struts2 + Spring + Hibernate为Web应用程序创建一个Login模块，如果用户希望通过该站点导航，我想强制用户登录。

我有一个LoginInterceptor

public class LoginInterceptor implements Interceptor {

public LoginInterceptor() {
}

public void destroy() {
    System.out.println("LoginInterceptor destroy() is called...");
}

public void init() {
    System.out.println("LoginInterceptor init() is called...");
}

public String intercept(ActionInvocation actionInvocation) throws Exception {

    final ActionContext context = actionInvocation.getInvocationContext();
    HttpServletRequest request = (HttpServletRequest) context.get(ServletActionContext.HTTP_REQUEST);
    HttpSession session = request.getSession(true);

    //Code

    try {

        Users userObj = (Users) session.getAttribute("userObj");

        if (userObj == null) {
            System.out.println("if from LoginInterceptor");
            return "noLogin";
        }

    } catch (Exception e) {
        Logger.getLogger(LoginInterceptor.class.getName()).log(Level.INFO, "message", e);
    }

    //Invoke action
    String result = actionInvocation.invoke();

    return result;
}
}

我的struts.xml

 <interceptors>

        <interceptor name="myLocale" class="com.deveto.struts.interceptors.LocaleInterceptor"/>
        <interceptor name="login" class="com.deveto.struts.interceptors.LoginInterceptor"/>
        <interceptor name="access" class="com.deveto.struts.interceptors.AccessInterceptor"/>

        <interceptor-stack name="defaultStack">
            <interceptor-ref name="myLocale"/>
            <interceptor-ref name="login"/>
            <interceptor-ref name="exception"/>
            <interceptor-ref name="alias"/>
            <interceptor-ref name="prepare"/>
            <interceptor-ref name="i18n"/>
            <interceptor-ref name="chain"/>
            <interceptor-ref name="debugging"/>
            <interceptor-ref name="profiling"/>
            <interceptor-ref name="fileUpload"/>
            <interceptor-ref name="checkbox"/>
            <interceptor-ref name="params">
                <param name="excludeParams">dojo\..*</param>
            </interceptor-ref>
            <interceptor-ref name="conversionError"/>
            <interceptor-ref name="validation">
                <param name="excludeMethods">input,back,cancel,browse</param>
            </interceptor-ref>
            <interceptor-ref name="workflow">
                <param name="excludeMethods">input,back,cancel,browse</param>
            </interceptor-ref>
        </interceptor-stack>
    </interceptors>

    <default-interceptor-ref name="defaultStack"/>

    <global-results>
        <result name="noLogin" type="redirectAction">show-login</result>
        <result name="noAccess" type="redirectAction">access-required</result>
    </global-results>

    <action
        name="show-login"
        class="com.deveto.struts.actions.UsersAction" >
        <interceptor-ref name="defaultStack"/>
        <result name="success" type="tiles">tiles.login</result>
    </action>

但是当我运行项目时，我有一个溢出异常，在堆栈跟踪中我什么都没有，但是Mozilla告诉我：

      The page isn't redirecting properly  

      Firefox has detected that the server is redirecting the request for this address in a way that will never complete.

和

  System.out.println("if from LoginInterceptor");

重复几次。我不明白为什么，但我有更多的拦截器，除此之外还有其他好处。

Answer 1

当您的LoginInterceptor返回其noLogin结果时，它会重定向并再次被LoginInterceptor拦截，从而将其转换为无限重定向循环。

因此，您必须排除show-login被LoginInterceptor拦截，例如。

定义两个拦截器堆栈，并将设置为LoginInterceptor 作为默认值：

<interceptor-stack name="defaultStack">
  ... same as in your question ...
</interceptor-stack>

<interceptor-stack name="noLoginStack">
  ... same as in your question but *without* the LoginInterceptor ...
</interceptor-stack>

<default-interceptor-ref name="defaultStack"/>

然后仅针对show-login操作，使用 noLoginStack ：

<action name="show-login"
    ... >
    <interceptor-ref name="noLoginStack"/>
    ...
</action>

Answer 2

为什么不直接使用Spring Security呢？配置非常简单，它与Struts2完美配合。从那里开始：http://static.springsource.org/spring-security/site/tutorial.html或那里：http://www.mularien.com/blog/2008/07/07/5-minute-guide-to-spring-security/

Struts2拦截器溢出异常

2 个答案: