使用Play框架在Scala中解析Json字符串 [英] Parsing a Json String in Scala using Play framework

问题描述

我开始尝试Scala和Play来解析Json数据，并在 https://www.playframework.com/documentation/2.3.9/ScalaJson . 现在，当我尝试运行给出的示例代码时:

I started trying Scala and Play to parse through Json data, and was following the tutorial at https://www.playframework.com/documentation/2.3.9/ScalaJson. Now, when I try to run the sample code given there which is:

val json: JsValue = Json.parse("""{
  "name" : "Watership Down",
  "location" : {
    "lat" : 51.235685,
    "long" : -1.309197
  },
  "residents" : [ {
    "name" : "Fiver",
    "age" : 4,
    "role" : null
  }, {
    "name" : "Bigwig",
    "age" : 6,
    "role" : "Owsla"
  } ]
}
""")

val lat = json \ "location" \ "lat"

我收到以下错误:

java.lang.NoSuchMethodError: play.api.libs.json.JsValue.$bslash(Ljava/lang/String;)Lplay/api/libs/json/JsValue;

我做错了什么?我正在使用Scala 2.10和Play 2.3.9.

What am I doing wrong? I'm using Scala 2.10 and Play 2.3.9.

谢谢.

推荐答案

在Play 2.4.x中，JsLookupResult表示特定Json路径(实际的Json节点或未定义)上的值. JsLookupResult具有两个子类:分别为JsDefined和JsUndefined.

In Play 2.4.x, JsLookupResult represents the value at a particular Json path, either an actual Json node or undefined. JsLookupResult has two subclasses: JsDefined and JsUndefined respectively.

您可以按以下方式修改代码:

You can modify your code as the following:

val name: JsLookupResult = json \ "user" \ "name"

name match {
  case JsDefined(v) => println(s"name = ${v.toString}")
  case undefined: JsUndefined => println(undefined.validationError)
}

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