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"
I get the following error:
java.lang.NoSuchMethodError: play.api.libs.json.JsValue.$bslash(Ljava/lang/String;)Lplay/api/libs/json/JsValue;
What am I doing wrong? I'm using Scala 2.10 and Play 2.3.9.
Thanks.
The JSON.parse() method parses a JSON string, constructing the JavaScript value or object described by the string. An optional reviver function can be provided to perform a transformation on the resulting object before it is returned.
Argonaut is a great library. It's by far the best JSON library for Scala, and the best JSON library on the JVM. If you're doing anything with JSON in Scala, you should be using Argonaut. circe is a fork of Argonaut with a few important differences.
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.
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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