问题
As per the question, how to encode Map[String, MyCaseClass] into Seq[String, String] using circe?
Model:
case class MyCaseClass(name: String, enabled: Boolean)
case class Parent(parentName: String,
collection: Map[String, MyCaseClass])
Parent(
"parent-name",
Map(
"external-name-a", MyCaseClass("internal-name-a", true),
"external-name-b", MyCaseClass("internal-name-b", false)
)
)
I'd like to encode this into a:
Seq[name = <map key>, enabled = <boolean value from MyCaseClass>]
For example:
{ ...
collection: [
{
name: "external-name-a",
enabled: true
},
{
name: "external-name-a",
enabled: false
}
]
...
}
I got as far as the following, just unsure exactly how to do what I want
object Parent {
implicit val encodeParent: Encoder[Parent] = (parent: Parent) => {
Json.obj(
("name", parent.name.asJson),
("collection", parent.collection.asJson),
)
}
implicit val encodeCollection: Encoder[Map[String, MyCaseClass]] = (collection: Map[String, MyCaseClass]) => {
//collection.toList.map((externalName: String, myCaseClass: MyCaseClass) => (externalName, myCaseClass.enabled)).asJson
}
}
回答1:
This works:
implicit val encodeCollection: Encoder[Map[String, MyCaseClass]] = (collection: Map[String, MyCaseClass]) => {
collection.toList.
map(collection => (collection._1, collection._2.enabled)).
map(collection => Json.obj(
("name", collection._1.asJson),
("enabled", collection._2.asJson)
)).asJson
}
来源:https://stackoverflow.com/questions/51378340/encode-mapstring-mycaseclass-into-seqstring-string-using-circe