Grails Spring Security获取当前页面的角色

我想知道是否有人知道一个优雅的方式来获取有权访问当前页面的Spring安全插件中的所有角色。Grails Spring Security获取当前页面的角色

我使用的是spring security，它被配置为使用RequestMap域对象。

我的应用程序的权限非常复杂，所以我想在每个页面的底部制作一个标签，以显示需要使用该页面的角色。

我在做一个查询请求地图，但我想确保我匹配url的方式与插件的方式相同。

理想情况下，我根本不需要运行查询。

Grails的2.2.1版本的Spring Security插件版本1.2.7.3

在此先感谢

来源

2014-02-27 davelt

我得到这个通过添加下列两类，以我的src/java的工作。

1类

import org.springframework.security.access.ConfigAttribute; 
import org.springframework.security.web.FilterInvocation; 
import org.springframework.security.web.access.intercept.FilterInvocationSecurityMetadataSource; 

import javax.servlet.http.HttpServletRequest; 
import java.util.Collection; 

public class MyFilterInvocationSecurityMetadataSource implements FilterInvocationSecurityMetadataSource { 

FilterInvocationSecurityMetadataSource oldBean; 

@Override 
public Collection<ConfigAttribute> getAttributes(Object o) throws IllegalArgumentException { 
    FilterInvocation filterInvocation = (FilterInvocation) o; 
    HttpServletRequest request = filterInvocation.getHttpRequest(); 
    request.setAttribute("PAGEROLES", oldBean.getAttributes(filterInvocation)); 

    return oldBean.getAttributes(o); 
} 

@Override 
public Collection<ConfigAttribute> getAllConfigAttributes() { 
    return oldBean.getAllConfigAttributes(); 
} 

@Override 
public boolean supports(Class<?> aClass) { 
    return FilterInvocation.class.isAssignableFrom(aClass); 
} 

public Object getOldBean() { return oldBean; } 
public void setOldBean(FilterInvocationSecurityMetadataSource oldBean) { this.oldBean = oldBean; } 
}

2类

import org.springframework.beans.BeansException; 
import org.springframework.beans.factory.BeanFactory; 
import org.springframework.beans.factory.BeanFactoryAware; 
import org.springframework.beans.factory.config.BeanPostProcessor; 
import org.springframework.beans.factory.config.ConfigurableListableBeanFactory; 
import org.springframework.security.web.access.intercept.FilterInvocationSecurityMetadataSource; 

public class FilterSecurityMDSExtractor implements BeanPostProcessor, BeanFactoryAware { 
    private ConfigurableListableBeanFactory bf; 
    private FilterInvocationSecurityMetadataSource metadataSource = new MyFilterInvocationSecurityMetadataSource(); 

    public Object postProcessBeforeInitialization(Object bean, String beanName) throws BeansException { 
     if (bean instanceof FilterInvocationSecurityMetadataSource) { 
      ((MyFilterInvocationSecurityMetadataSource) metadataSource).setOldBean((FilterInvocationSecurityMetadataSource) bean); 
      return metadataSource; 
     } 
     return bean; 
    } 

    public Object postProcessAfterInitialization(Object bean, String beanName) throws BeansException { 
     return bean; 
    } 

    public void setBeanFactory(BeanFactory beanFactory) throws BeansException { 
     this.bf = (ConfigurableListableBeanFactory)beanFactory; 
    } 
}

我然后加入下列到resources.groovy

beans = { 
     filterSecurityMDSExtractor(FilterSecurityMDSExtractor) 
}

基本上我馅用户角色进入请求

request.setAttribute("PAGEROLES", oldBean.getAttributes(filterInvocation));

那么我需要做的就是调用下面

request.getAttribute("PAGEROLES");

获得角色退了出来。我通过从Stackoverflow上的其他好帖子窃取我的解决方案。其他人可能会有更好的解决方案，但到目前为止，这对我来说很有用。

来源

2014-02-28 18:31:08 davelt

Grails Spring Security获取当前页面的角色

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