How to hide a parameter in swagger?

Question

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.

Answer 1

Hope this helps.

For Fields

@ApiModelProperty(required = false, hidden = true)
private String hiddenProperty

For Apis

@ApiIgnore
public class MyApi {}

For Parameters

public void getApi(@ApiIgnore String param){}

@ApiModelProperty(hidden="true")
public String paramInsideClass

Answer 2

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
  }
}

And then enable it in 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>