Question

对于未答复的请求，我想隐藏此字段（模型架构）。

我的要求

@ApiOperation(value = "Create node")
@ApiImplicitParams({
        @ApiImplicitParam(paramType = "body", required = true)
})
public Result insert()

我不想在@ApiOperation中显示响应属性。它有可能吗？

谢谢！

Answer 1

我不确定你在问什么。但是，如果您尝试在JSON响应中隐藏模型中的特定字段，请尝试使用fastxml的jackson-annotations模块中的$ ./test $ echo $? 0 $注释。只需将注释添加到您在响应中尝试避免的字段中。

Answer 2

即使您没有说，我相信您在https://github.com/swagger-api/swagger-play/pull/76#issuecomment-224287765发布的帖子表明您正在使用Play Framework。我相信目前招摇的“无效”结果已被打破（参考：https://github.com/swagger-api/swagger-play/issues/89）。我在Scala中执行此操作的一种方法是在responseReference = "void"中提供@ApiOperation并执行以下操作作为我的Scala控制器，以显示Swagger规范以覆盖我的更改：

package controllers

import controllers.SwaggerBaseApiController
import io.swagger.models.auth.{ApiKeyAuthDefinition, BasicAuthDefinition, In}
import io.swagger.models.properties.RefProperty
import io.swagger.models.{Info, Response}
import play.api.mvc.Action

import scala.collection.JavaConversions._

object Swagger extends SwaggerBaseApiController {

  def json = Action { implicit request =>
    val swagger = getResourceListing(request.host)
    // We need to modify this if it doesn't contain our security definitions yet, but we have to do it atomically
    // This should be fast enough that this synchronization is not too bad
    swagger.synchronized {
      if (!somethingThreadSafeToShowYouveChangedItAlready) fixSwagger(swagger)
    }
    // We trust that the above code only changes a swagger instance once therefore we don't need to
    // synchronize the json marshalling because it should not change beneath it
    returnValue(request, toJsonString(swagger))
  }

  private[this] def fixSwagger(swagger: io.swagger.models.Swagger): Unit = {
    // Omitted some of my other changes...

    swagger.getPaths.values.foreach { value =>

      value.getOperations.foreach { oper =>
        // Omitted some of my other chabnges

        // Any responses that are void need to be simple void
        oper.getResponses.values.foreach { resp =>
          resp.getSchema() match {
            case schema: RefProperty if schema.get$ref() == "#/definitions/void" => resp.setSchema(null)
            case _ => ()
          }
        }
      }
    }
  }
}

Answer 3

隐藏所有控制器API

  @ApiIgnore

隐藏所选属性

@ApiModelProperty(required = false, hidden = true)

示例：可见

@ApiModelProperty(
        access = "public",
        name = "amount",
        example = "123.45",
        value = "the amount - in this example without currency.")
public String getAmount() {
    return amount;
}

示例：隐藏

@ApiModelProperty(
       required = false,
       hidden = true
    )
public String getAmount() {
    return amount;
}

如何在Swagger中隐藏模式模型？

3 个答案: