如何在招摇中隐藏参数? [英] How to hide a parameter in swagger?

问题描述

是否有人成功地从生成的文档中隐藏了参数?我发现一个问题 here，但使用 @ApiParam(access="@HeaderParam 之前的 internal", required=false) 似乎不起作用.

did anyone succeed to hide a parameter from generated documentation? I found an issue here, but using @ApiParam(access="internal", required=false) before @HeaderParam did not seem to work.

推荐答案

好的，看看单元测试有帮助.首先你需要定义一个过滤器:

Ok, looking at the unit tests helped. First you need to define a filter:

import com.wordnik.swagger.core.filter.SwaggerSpecFilter
import com.wordnik.swagger.model.{Parameter, ApiDescription, Operation}
import java.util

class MySwaggerSpecFilter extends SwaggerSpecFilter{
  override def isOperationAllowed(operation: Operation, api: ApiDescription, params: util.Map[String, util.List[String]], cookies: util.Map[String, String], headers: util.Map[String, util.List[String]]): Boolean = true

  override def isParamAllowed(parameter: Parameter, operation: Operation, api: ApiDescription, params: util.Map[String, util.List[String]], cookies: util.Map[String, String], headers: util.Map[String, util.List[String]]): Boolean = {
    if(parameter.paramAccess == Some("internal")) false
    else true
  }
}

然后在web.xml

    <servlet>
        <servlet-name>DefaultJaxrsConfig</servlet-name>
        <servlet-class>com.wordnik.swagger.jaxrs.config.DefaultJaxrsConfig</servlet-class>
        ...
        <init-param>
            <param-name>swagger.filter</param-name>
            <param-value>com.example.MySwaggerSpecFilter</param-value>
        </init-param>
    </servlet>

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