scala Play 框架 - 向 JSON 对象添加字段
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Play Framework - add a field to JSON object
提问by Pawe? Kozikowski
I have a problem with adding a field to Json object in Play Framework using Scala:
我在使用 Scala 向 Play Framework 中的 Json 对象添加字段时遇到问题：
I have a case class containing data. For example:
我有一个包含数据的案例类。例如：
case class ClassA(a:Int,b:Int)
and I am able to create a Json object using Json Writes:
我可以使用 Json Writes 创建一个 Json 对象：
val classAObject = ClassA(1,2)
implicit val classAWrites= Json.writes[ClassA]
val jsonObject = Json.toJson(classAObject)
and the Json would look like:
和 Json 看起来像：
{ a:1, b:2 }
Let's suppose I would like to add an additional 'c' field to the Json object. Result:
假设我想向 Json 对象添加一个额外的“c”字段。结果：
{ a:1, b:2, c:3 }
How do I do that without creating a new case class or creating my Json object myself using Json.obj? I am looking for something like:
如何在不创建新案例类或使用 Json.obj 自己创建 Json 对象的情况下做到这一点？我正在寻找类似的东西：
jsonObject.merge({c:3})
Any help appreciated!
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回答by Travis Brown
JsObjecthas a +method that allows you to add fields to an object, but unfortunately your jsonObjectis statically typed as a JsValue, not a JsObject. You can get around this in a couple of ways. The first is to use as:
JsObject有一个+方法可以让您向对象添加字段，但不幸的jsonObject是，您的静态类型为 a JsValue，而不是 a JsObject。您可以通过多种方式解决此问题。第一种是使用as：
scala> jsonObject.as[JsObject] + ("c" -> Json.toJson(3))
res0: play.api.libs.json.JsObject = {"a":1,"b":2,"c":3}
With asyou're essentially downcasting—you're telling the compiler, "you only know that this is a JsValue, but believe me, it's also a JsObject". This is safe in this case, but it's not a good idea. A more principled approach is to use the OWritesdirectly:
随着as你基本上向下转换-你告诉编译器，“你只知道这是一个JsValue，但相信我，这也是一个JsObject”。在这种情况下这是安全的，但这不是一个好主意。一个更有原则的方法是OWrites直接使用：
scala> val jsonObject = classAWrites.writes(classAObject)
jsonObject: play.api.libs.json.JsObject = {"a":1,"b":2}
scala> jsonObject + ("c" -> Json.toJson(3))
res1: play.api.libs.json.JsObject = {"a":1,"b":2,"c":3}
Maybe someday the Jsonobject will have a toJsonObjectmethod that will require a OWritesinstance and this overly explicit approach won't be necessary.
也许有一天该Json对象会有一个toJsonObject需要OWrites实例的方法，而这种过于明确的方法将是不必要的。
回答by Pawe? Kozikowski
I found a solution myself. In fact the JsValue, which is the return type of Json.toJson has no such method, but the JsObject (http://www.playframework.com/documentation/2.2.x/api/scala/index.html#play.api.libs.json.JsObject) does, so the solution is:
我自己找到了解决方案。实际上，Json.toJson 的返回类型 JsValue 没有这样的方法，但是 JsObject ( http://www.playframework.com/documentation/2.2.x/api/scala/index.html#play.api .libs.json.JsObject) 确实如此，所以解决方案是：
val jsonObject = Json.toJson(classAObject).as[JsObject]
jsonObject + ("c", JsNumber(3))
I hope someone will find this useful :)
我希望有人会发现这很有用:)
回答by Vishnu Kumar
simpler way is to use argoanut (http://argonaut.io/)
更简单的方法是使用 argoanut ( http://argonaut.io/)
var jField : Json.JsonField = "myfield" //Json.JsonField is of type String
obj1.asJson.->:(jField, obj2.asJson) // adds a field to obj1.asJson
here obj1.asJsoncreates a JSON object
and obj2is the object to be added to the json created by obj1.asJson
这里obj1.asJson创建了一个 JSON 对象，并且obj2是要添加到由创建的 json 中的对象obj1.asJson
