如何在招摇中隐藏参数? [英] How to hide a parameter in swagger?
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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="
似乎不起作用.@HeaderParam
之前的 internal", required=false)
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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