Question

我试图编写自己的Swagger API过滤器来隐藏某些内容，使其不会出现在API中。我想我对应该做什么有很好的了解，但我必须在细节中遗漏一些东西。

我使用的是Swagger Core 1.3.10和Spring 3.1。这是我现在拥有的一些内容：

SwaggerApiAuthorizationFilter

public class SwaggerApiAuthorizationFilter implements SwaggerSpecFilter {

    private final static Logger logger = LoggerFactory.getLogger(SwaggerApiAuthorizationFilter.class);

    @Override
    public boolean isOperationAllowed(Operation operation, ApiDescription api, Map<String, List<String>> params, 
            Map<String, String> cookies, Map<String, List<String>> headers) {

        return true;
    }

    @Override
    public boolean isParamAllowed(Parameter parameter, Operation operation, ApiDescription api, Map<String, List<String>> params, 
            Map<String, String> cookies, Map<String, List<String>> headers) {

        // do not allow the documentation to be generated on parameters that have their access set to "internal"
        if (parameter.paramAccess().isDefined() && parameter.paramAccess().get().equalsIgnoreCase("internal")) {

            logger.debug("The following parameter has been hidden from the Swagger API documentation: " + parameter.name());

            return false;

        } else {

            return true;
        }
    }

}

SwaggerServlet

public class SwaggerServlet extends HttpServlet {

    private static final long serialVersionUID = 1L;
    private final static Logger logger = LoggerFactory.getLogger(SwaggerServlet.class);

    BeanConfig beanConfig;
    SwaggerApiAuthorizationFilter swaggerApiAuthorizationFilter;

    public void setBeanConfig(BeanConfig beanConfig) {
        this.beanConfig = beanConfig;
    }

    public void setSwaggerApiAuthorizationFilter(SwaggerApiAuthorizationFilter swaggerApiAuthorizationFilter) {
        this.swaggerApiAuthorizationFilter = swaggerApiAuthorizationFilter;
    }

    @Override
    public void init(ServletConfig servletConfig) throws ServletException {
        try {

            super.init(servletConfig);

            beanConfig.setBasePath("/mbl/services");
            beanConfig.setVersion("1.0");
            beanConfig.setResourcePackage("com.whatever.resources");
            beanConfig.setScan(true);

            FilterFactory.setFilter(swaggerApiAuthorizationFilter);

            logger.debug("The Swagger servlet has been initialized");

        } catch (Exception e) {

            e.printStackTrace();
        }
    }
}

弹簧配置文件

<!-- Swagger Configuration and Providers -->
<bean id="beanConfig" class="com.wordnik.swagger.jaxrs.config.BeanConfig">
    <property name="title" value="Java API"/>
    <property name="version" value="1.0" />
    <property name="basePath" value="/mbl/services"/>
    <property name="resourcePackage" value="com.whatever.resources"/>
    <property name="scan" value="true"/>
</bean>

<bean id="swaggerApiAuthorizationFilter" class="com.whatever.util.SwaggerApiAuthorizationFilter" />

的web.xml

<!-- Enabling Swagger servlet -->
<servlet>
    <servlet-name>Swagger Servlet</servlet-name>
    <servlet-class>com.whatever.web.SwaggerServlet</servlet-class>
        <init-param>
            <param-name>swagger.filter</param-name>
            <param-value>com.whatever.util.SwaggerApiAuthorizationFilter</param-value>
        </init-param>
    <load-on-startup>-1</load-on-startup> 
</servlet>
<servlet-mapping>
    <servlet-name>Swagger Servlet</servlet-name>
    <url-pattern>/api-docs</url-pattern>
</servlet-mapping>

资源类

    public Response getUserInfo( 
    @Context HttpHeaders headers, 
    @ApiParam(name="brand", access="internal") @CookieParam(value = "brand") String brand) {

看来我的SwaggerApiAuthorizationFilter甚至没有被调用。我的猜测是我的配置在web.xml或SwaggerServlet中出错了。有没有人看到上面看起来不正确或有没有任何想法？

Answer 1

我刚刚成功使用了过滤器。我有问题使过滤功能正常。问题是找到我的过滤器的实现类是一个问题。当我将包改为swagger类可以访问的地方时，所有内容都正确无误。接下来，我只需要在我的过滤器类上实现我的逻辑。我的过滤器类也实现了SwaggerSpecFilter接口。最好的问候。

我正在使用DefaultJaxrsConfig：

<servlet>
  <servlet-name>Jersey2Config</servlet-name>
    <servlet-class>io.swagger.jaxrs.config.DefaultJaxrsConfig</servlet-class>
    <init-param>
        <param-name>api.version</param-name>
        <param-value>1.0.0</param-value>
    </init-param>
    <init-param>
        <param-name>swagger.api.basepath</param-name>
        <param-value>/my-service/service/</param-value>
    </init-param>
    <init-param>
        <param-name>swagger.filter</param-name>
        <param-value>mypackage.SwaggerFilter</param-value>
    </init-param>
    <load-on-startup>3</load-on-startup>
</servlet>

Answer 2

我只是在猜测，但我认为你需要实现Filter以及SwaggerSpecFilter接口，而不是扩展HttpServlet（看看它没有出现SwaggerSpecFilter扩展过滤器的源代码，然后将其配置为过滤器）在您的web.xml而不是servlet中。它会是这样的：

<filter> <filter-name>Swagger Filter</filter-name> <filter-class>com.whatever.util.SwaggerApiAuthorizationFilter</filter-class> </filter> <filter-mapping> <filter-name>Swagger Filter</filter-name> <url-pattern>/api-docs</url-pattern> </filter-mapping>

同样，不确定这是否会起作用，但只是猜测我的头脑。

Swagger - 扩展SwaggerSpecFilter

2 个答案: