如何从 Scala 中的响应 JSON 中删除 List() 和转义字符

Question

我有以下函数，它接收 JSON 输入并使用 "com.eclipsesource" %% "play-json-schema-validator" % "0.6.2" 库根据 JSON 模式对其进行验证。当我收到无效的 JSON 时，一切正常，我尝试将所有违规行为收集到 List 中，然后将该列表与响应 JSON 一起返回。但是，我的 List 是用 List() 编码的，并且还有转义字符。我想让响应 JSON 看起来像这样：

{
  "transactionID": "123",
  "status": "error",
  "description": "Invalid Request Received",
  "violations": ["Wrong type. Expected integer, was string.", "Property action missing"]
}

而不是这个：（这是我现在得到的）

{
  "transactionID": "\"123\"",
  "status": "error",
  "description": "Invalid Request Received",
  "violations": "List(\"Wrong type. Expected integer, was string.\", \"Property action missing\")"
}

这是 JSON 验证的实际功能

def validateRequest(json: JsValue): Result = {

    {
      val logger = LoggerFactory.getLogger("superman")
      val jsonSchema = Source.fromFile(play.api.Play.getFile("conf/schema.json")).getLines.mkString
      val transactionID = (json \ "transactionID").get
      val result: VA[JsValue] = SchemaValidator.validate(Json.fromJson[SchemaType](
        Json.parse(jsonSchema.stripMargin)).get, json)

      result.fold(
        invalid = { errors =>

          var violatesList = List[String]()
          var invalidError = Map("transactionID" -> transactionID.toString(), "status" -> "error", "description" -> "Invalid Request Received")
          for (msg <- (errors.toJson \\ "msgs"))
            violatesList = (msg(0).get).toString() :: violatesList
          invalidError += ("violations" -> (violatesList.toString()))
          //TODO: Make this parsable JSON list
          val errorResponse = Json.toJson(invalidError)
          logger.error("""Message="Invalid Request Received" for transactionID=""" + transactionID.toString() + "errorResponse:" + errorResponse)
          BadRequest(errorResponse)

        },

        valid = {
          post =>
            db.writeDocument(json)
            val successResponse = Json.obj("transactionID" -> transactionID.toString, "status" -> "OK", "message" -> ("Valid Request Received"))
            logger.info("""Message="Valid Request Received" for transactionID=""" + transactionID.toString() + "jsonResponse:" + successResponse)
            Ok(successResponse)
        }
      )
    }

  }

更新 1

我在使用 Json.obj() 后得到了这个

{
  "transactionID": "\"123\"",
  "status": "error",
  "description": "Invalid Request Received",
  "violations": [
    "\"Wrong type. Expected integer, was string.\"",
    "\"Property action missing\""
  ]
}

Answer 1

我通过修改这一行删除了转义字符：

violatesList = (msg(0).get).toString() :: violatesList

到：

violatesList = (msg(0).get).as[String] :: violatesList

Answer 2

您想要的是一个 JSON 数组，但通过在列表中调用 .toString()，您实际上是在传递一个字符串。 Play 有一个将 Lists 转换为 JSON 数组的隐式序列化程序，因此您实际上只需要做的比已经做的少 - 您可以从 violatesList.toString() 中删除 toString() 部分。

此外，不要为您的 JSON 创建映射然后将其转换为 JSON，您可以使用 Json.obj，语法非常相似：

val invalidError = Json.obj("transactionID" -> transactionID, "status" -> "error", "description" -> "Invalid Request Received")
for (msg <- (errors.toJson \\ "msgs"))
  violatesList = (msg(0).get) :: violatesList
val errorResponse = invalidError ++ Json.obj("violations" -> violatesList)

关于你的转义引号，我想这是因为transactionID 和msgs 是JsStrings，所以当你用toString() 转换它们时，引号包括在内。只需在各处删除toString 就可以了。