Having a jString : JString
value holding an "abc"
string inside I get "JString(abc)" : String
if I call jString.toString
. How do I get "abc" : String
instead?
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21

Ivan
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1I've found the solution to use `jString.values : String` but I feel uncertain if it's correct - why is it called `values` (not a `value`) if there's just a `String`? – Ivan Oct 16 '11 at 03:19
4 Answers
19
To extract a value from JValue you can use any method described here: What is the most straightforward way to parse JSON in Scala?
For instance:
json.extract[String]
You can use 'render' function to convert any JValue to printable format. Then either 'pretty' or 'compact' will convert that to a String.
compact(render(json))
or
pretty(render(json))
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I don't want to render JSON. I want the opposite - to extract a string value of a property of an object serialized in JSON and I've already isolated that only field into a separate JString, containing nothing but the value I need. – Ivan Oct 16 '11 at 09:18
7
I believe the best way is to use match:
val x = ... (whatever, maybe it's a JString)
x match {
case JString(s) => do something with s
case _ => oops, something went wrong
}

Vlad Patryshev
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7
val jstring=JString("abc")
implicit val formats = net.liftweb.json.DefaultFormats
System.out.println(jstring.extract[String])

Win Myo Htet
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1
This was asked a while ago, but I wanted a simple one-line helper that would get my string for me in the context of an expression, so I wrote this little thing inside of an object called Get:
object Get {
def string(value: JValue): String = {
val JString(result) = value
result
}
...
}
This way I can just do, e.g., val myString = Get.string(jsonStringValue)

Necro
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