1

我想解析一个内容为 json 格式的文件。我想从文件中提取一些属性（名称、DataType、Nullable）来动态创建一些列名。我已经浏览了一些示例，但其中大多数都使用案例类，但我的问题是每次收到文件时可能有不同的内容。

我尝试使用ujson库来解析文件，但我无法理解如何正确使用它。

object JsonTest {
  def main(args: Array[String]): Unit = {

    val source = scala.io.Source.fromFile("C:\\Users\\ktngme\\Desktop\\ass\\file.txt")
    println(source)
    val input = try source.mkString finally source.close()
    println(input)

    val data = ujson.read(input)
    data("name") = data("name").str.reverse
    val updated = data.render()
  }
}

文件示例的内容：

{
"Organization": {
"project": {
"name": "POC 4PL",
"description": "Implementation of orderbook"
},
"Entities": [
{
"name": "Shipments",
"Type": "Fact",
"Attributes": [
{
"name": "Shipment_Details",
"DataType": "StringType",
"Nullable": "true"
},
{
"name": "Shipment_ID",
"DataType": "StringType",
"Nullable": "true"
},
{
"name": "View_Cost",
"DataType": "StringType",
"Nullable": "true"
}
],
"ADLS_Location": "/mnt/mns/adls/raw/poc/orderbook/"
}
]
}
}

预期输出：

StructType(
Array(StructField("Shipment_Details",StringType,true),
StructField("Shipment_ID",DateType,true),   
StructField("View_Cost",DateType,true)))

StructType 需要以编程方式添加到预期的输出中。

4

2 回答 2

0

这取决于您是否希望它完全动态，这里有一些选项：

如果您只想阅读一个字段，您可以执行以下操作： 

import upickle.default._

val source = scala.io.Source.fromFile("C:\\Users\\ktngme\\Desktop\\ass\\file.txt")
val input = try source.mkString finally source.close()
val json = ujson.read(input)

println(json("Organization")("project")("name"))

输出将是："POC 4PL"

如果您只希望属性与类型一起使用，您可以执行以下操作： 

import upickle.default.{macroRW, ReadWriter => RW}
import upickle.default._

val source = scala.io.Source.fromFile("C:\\Users\\ktngme\\Desktop\\ass\\file.txt")
val input = try source.mkString finally source.close()
val json = ujson.read(input)
val entitiesArray = json("Organization")("Entities")(0)("Attributes")
println(read[Seq[StructField]](entitiesArray))

case class StructField(name: String, DataType: String, Nullable: String)
object StructField{
  implicit val rw: RW[StructField] = macroRW
}

输出将是：List(StructField(Shipment_Details,StringType,true), StructField(Shipment_ID,StringType,true), StructField(View_Cost,StringType,true))

另一种选择是使用不同的库来进行类映射。如果您使用 Google Protobuf Struct和JsonFormat它可以是 2-liner： 

import com.google.protobuf.Struct
import com.google.protobuf.util.JsonFormat

val source = scala.io.Source.fromFile("C:\\Users\\ktngme\\Desktop\\ass\\file.txt")
val input = try source.mkString finally source.close()

JsonFormat.parser().merge(input, builder)
println(builder.build())

输出将是：fields { key: "Organization" value { struct_value { fields { key: "project" value { struct_value { fields { key: "name" value { string_value: "POC 4PL" } } fields { key: "description" value { string_value: "Implementation of orderbook" } } } } } fields { key: "Entities" value { list_value { values { struct_value { fields { key: "name" value { string_value: "Shipments" } }...

于 2019-07-21T14:55:34.443 回答
0

尝试使用 Playframework 的 Json 工具 - https://www.playframework.com/documentation/2.7.x/ScalaJson

这是您的问题的解决方案- \ 将您的 json 放在文本文件中

    val fil_path = "C:\\TestData\\Config\\Conf.txt"
    val conf_source = scala.io.Source.fromFile(fil_path)
    lazy val json_str = try conf_source.mkString finally conf_source.close()
    val conf_json: JsValue = Json.parse(json_str)
    val all_entities: JsArray = (conf_json \ "Organization" \ "Entities").get.asInstanceOf[JsArray]
    val shipments: JsValue = all_entities.value.filter(e => e.\("name").as[String] == "Shipments").head
    val shipments_attributes: IndexedSeq[JsValue] = shipments.\("Attributes").get.asInstanceOf[JsArray].value
    val shipments_schema: StructType = StructType(shipments_attributes.map(a => Tuple3(a.\("name").as[String], a.\("DataType").as[String], a.\("Nullable").as[String]))
      .map(x => StructField(x._1, StrtoDatatype(x._2), x._3.toBoolean)))
    shipments_schema.fields.foreach(println)

输出是 -

StructField(Shipment_Details,StringType,true)
StructField(Shipment_ID,StringType,true)
StructField(View_Cost,StringType,true)
于 2019-07-29T05:43:02.817 回答