2

I'm using liftweb to parse JSON from String in scala, some of record have 3 field 

val a = {"name": "Alice", "age": 21, "job": "nurse"}

but some other have only 2 field 

val b = {"name": "Bob", "age": 30}

I created case class Person(name: String, age: Long, job: String) and when I call parse(a) it return value successfully, but when I call parse(b) it appear exception 

net.liftweb.json.MappingException: No usable value for algorithm
Did not find value which can be converted into java.lang.String
Loc Tran
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2 Answers2

3

If you make the parameter type job:String you are going to have issues since that would require the parameter to have a value - and in your example it doesn't.

I'll assume we want to make that an Option[String] and in the example below just add multiple constructors to match your parameters. Something like this should work:

case class Person(name: String, age: Long, job: Option[String]){
  def this(name: String, age: Long) = this(name, age, None)
}

If you had a default value, and wanted job to be a String just change the None to whatever you want by default.

After that, parsing as you did above should work for both cases.

jcern
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0

Use lift-json_2.11 version 2.6.3 to fix this issue. I also came across this issue with version 3.2.0

Sandeep Ballu
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