Given the following JSON:
{
"field1": "value1",
"field2": "",
"field3": "value3",
"field4": ""
}
How do I get two distinct JSONs, one containing the fields with value and another one containing the fields without value? Here below is how the final result should look like:
{
"field1": "value1",
"field3": "value3"
}
{
"field2": "",
"field4": ""
}
You have access to the JSON object's fields as a sequence of (String, JsValue) pairs and you can filter through them. You can filter out the ones with and without value and use the filtered sequences to construct new JsObject objects.
import play.api.libs.json._
val ls =
("field1", JsString("value1")) ::
("field2", JsString("")) ::
("field3", JsString("value3")) ::
("field4", JsString("")) ::
Nil
val js0 = new JsObject(ls)
def withoutValue(v: JsValue) = v match {
case JsNull => true
case JsString("") => true
case _ => false
}
val js1 = JsObject(js0.fields.filterNot(t => withoutValue(t._2)))
val js2 = JsObject(js0.fields.filter(t => withoutValue(t._2)))
You can improve @nietaki solution by using partition function.
import play.api.libs.json._
val ls =
("field1", JsString("value1")) ::
("field2", JsString("")) ::
("field3", JsString("value3")) ::
("field4", JsString("")) ::
Nil
val js0 = new JsObject(ls)
def withoutValue(v: JsValue) = v match {
case JsNull => true
case JsString("") => true
case _ => false
}
val (js1, js2) = js0.fields.partition(t => withoutValue(t._2))
JsObject(js1)
JsObject(js2)
You can also do it in a one liner:
val (js1, js2) = j.fields.partition(_._2 != JsString(""))
println(JsObject(js1))
println(JsObject(js2))
Code run at Scastie