Question

我是scala中的一个完整的新手，我想在这里做的只是使用scala解析并打印相同的JSON文件。我在编译时面临一个错误，我无法解决。在此先感谢任何帮助。 PFB scala代码，SBT文件，JSON文件和错误：

json_example.scala

import scala.io.Source
import play.api.libs.json._
import play.api.libs.json._


object Test extends App  {


val line :String = "Foo";

val filename = "users.json"
for (line <- Source.fromFile(filename).getLines().mkString) {
println(line);
val json: JsValue = Json.parse(line);

 }
 }

JSON文件（ users.json ）

{"users":[
    {"ID":"1","firstName":"John", "lastName":"Doe"},
    {"ID":"2","firstName":"Anna", "lastName":"Smith"},
    {"ID":"3","firstName":"Peter", "lastName":"Jones"}
    {"ID":"1","firstName":"Stewie", "lastName":"Doe"},
    {"ID":"2","firstName":"Chris", "lastName":"Smith"},
    {"ID":"3","firstName":"Louis", "lastName":"Jones"}
    {"ID":"2","firstName":"Brian", "lastName":"Smith"},
    {"ID":"3","firstName":"Meg", "lastName":"Jones"}

]}

SBT文件（simple.sbt）

lazy val root = (project in file(".")).

settings(
name := "JSON_GRAPHX",
version := "1.0",
scalaVersion := "2.10.1",

libraryDependencies ++= Seq("com.github.scala-incubator.io" %% "scala-io-file" % "0.4.2",
                       "com.typesafe.play" %% "play-json" % "2.3.4"),

                      resolvers += "Typesafe Repo" at "http://repo.typesafe.com/typesafe/releases/"


)

错误

[info] Set current project to JSON_GRAPHX (in build file:/F:/Graphx_app/JSON_GRAPHX/)
[info] Compiling 1 Scala source to F:\Graphx_app\JSON_GRAPHX\target\scala-2.10\classes...
[error] F:\Graphx_app\JSON_GRAPHX\json_example.scala:15: overloaded method value parse with alternatives:
[error]   (input: Array[Byte])play.api.libs.json.JsValue <and>
[error]   (input: String)play.api.libs.json.JsValue
[error]  cannot be applied to (Char)
[error]  val json: JsValue = Json.parse(line);
[error]                           ^
[error] one error found
[error] (compile:compileIncremental) Compilation failed
[error] Total time: 4 s, completed Aug 18, 2016 8:51:22 PM

Answer 1

要构建JSON对象，您可以直接在{ "localTrain": "T7", "TC": "2", "TimeSheet": [ ["01", "London", "BXP", "T7", "1632", "1640"], ["01", "London", "BXP", "T7", "1632", "1640"], ["02", "Shanghai", "QWE", "T7", "1200", "1240"], ["02", "Shanghai", "QWE", "T7", "1200", "1240"], ["03", "LosAngeles", "DFG", "T7", "1300", "1340"], ["03", "LosAngeles", "DFG", "T7", "1300", "1340"], ["04", "NewDelhi", "VGH", "T7", "1400", "1440"], ["04", "NewDelhi", "VGH", "T7", "1400", "1440"], ["05", "Sydney", "SAW", "T7", "1500", "1540"], ["04", "NewDelhi", "VGH", "T7", "1400", "1440"], ["06", "Tokyo", "SAT", "T7", "1600", "1640"], ["06", "Tokyo", "SAT", "T7", "1600", "1640"], ["07", "Seoul", "BBT", "T7", "1700", "1740"], ["07", "Seoul", "BBT", "T7", "1700", "1740"], ["08", "CapeTown", "OOP", "T7", "1800", "1840"], ["08", "CapeTown", "OOP", "T7", "1800", "1840"], ] }返回的parse中调用String方法。类似的东西：

mkString

通过做：

val json = Json.parse(Source.fromFile(filename).getLines().mkString)

您实际上正在迭代Json字符串的所有字符 - 这就是您收到for (line <- Source.fromFile(filename).getLines().mkString)方法无法应用于Char的错误的原因。

获得JSON对象后，可以将其缩小打印：

parse

或者您可以以可读格式打印：

println(Json.stringify(json))

使用Play在Scala（2.10.1）中解析JSON文件

1 个答案: