How to hide a parameter in swagger?

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

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>

If you are using io.swagger.v3, you should use @Parameter(hidden = true) as described on the migration guide here https://springdoc.org/migrating-from-springfox.html

With swagger-springmvc (https://github.com/springfox/springfox) at the moment there's no way to use SwaggerSpecFilter. But it respects @ApiIgnore annotation - it can be applied to method parameter which shouldn't appear in generated metadata.