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Java 如何捕获401403个错误并作为成功响应200_Java_Spring_Hibernate_Web_Oauth - Fatal编程技术网

Java 如何捕获401403个错误并作为成功响应200

Java 如何捕获401403个错误并作为成功响应200,java,spring,hibernate,web,oauth,Java,Spring,Hibernate,Web,Oauth,我在java spring项目中工作,使用restful web服务。出于安全原因,我添加了oAuth spring安全性。当访问令牌过期时,它会显示401网关错误。我需要捕获这些错误,并用json错误响应作为200Http成功状态。例如{“error”:“Unauthorized”}。如何配置以捕获401错误。请帮助我解决此问题。您应该使用。创建过滤器以拥有自定义实现并添加它。下面是文档中的示例 <bean id="filterChainProxy" class="org.springf

我在java spring项目中工作,使用restful web服务。出于安全原因,我添加了oAuth spring安全性。当访问令牌过期时,它会显示401网关错误。我需要捕获这些错误,并用json错误响应作为200Http成功状态。例如
{“error”:“Unauthorized”}
。如何配置以捕获401错误。请帮助我解决此问题。

您应该使用。创建过滤器以拥有自定义实现并添加它。下面是文档中的示例

<bean id="filterChainProxy" class="org.springframework.security.web.FilterChainProxy">
  <sec:filter-chain-map path-type="ant">
     <sec:filter-chain pattern="/webServices/**" filters="
           securityContextPersistenceFilterWithASCFalse,
           basicAuthenticationFilter,
           exceptionTranslationFilter,
           filterSecurityInterceptor" />
     <sec:filter-chain pattern="/**" filters="
           securityContextPersistenceFilterWithASCTrue,
           formLoginFilter,
           exceptionTranslationFilter,
           filterSecurityInterceptor" />
  </sec:filter-chain-map>
</bean>

您应该使用。创建过滤器以拥有自定义实现并添加它。下面是文档中的示例

<bean id="filterChainProxy" class="org.springframework.security.web.FilterChainProxy">
  <sec:filter-chain-map path-type="ant">
     <sec:filter-chain pattern="/webServices/**" filters="
           securityContextPersistenceFilterWithASCFalse,
           basicAuthenticationFilter,
           exceptionTranslationFilter,
           filterSecurityInterceptor" />
     <sec:filter-chain pattern="/**" filters="
           securityContextPersistenceFilterWithASCTrue,
           formLoginFilter,
           exceptionTranslationFilter,
           filterSecurityInterceptor" />
  </sec:filter-chain-map>
</bean>

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