CODEWITHSUNDEEP
stringjsonscalaliftlift-json
Ivan
Ivan

Reputation: 64207

How to convert a JSON JString value to an ordinary String in Lift?

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?

Upvotes: 21

Views: 17018

Answers (4)

Necro
Necro

Reputation: 571

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)

Upvotes: 1

Vlad Patryshev
Vlad Patryshev

Reputation: 1422

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
}

Upvotes: 7

Win Myo Htet
Win Myo Htet

Reputation: 5457

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

Upvotes: 7

Joni
Joni

Reputation: 2789

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))

Upvotes: 19

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