scala 使用lift-json将scala对象序列化为JSon字符串
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Serializing a scala object into a JSon String using lift-json
提问by Ali Salehi
I am wondering, would you please let me know how can I use lift-json to serialize a simple bean class into json string (I'm using v2.0-M1). I tried:
我想知道，请让我知道如何使用lift-json 将一个简单的bean 类序列化为json 字符串（我使用的是v2.0-M1）。我试过：
val r = JsonDSL.pretty(JsonAST.render(myBean))
and I am getting
我得到
[error] found : MyBean
[error] required: net.liftweb.json.JsonAST.JValue
Thanks, -A
谢谢
回答by Joni
You can "decompose" a case class into JSON and then render it. Example:
您可以将案例类“分解”为 JSON，然后呈现它。例子：
scala> import net.liftweb.json.JsonAST._
scala> import net.liftweb.json.Extraction._
scala> import net.liftweb.json.Printer._
scala> implicit val formats = net.liftweb.json.DefaultFormats
scala> case class MyBean(name: String, age: Int)
scala> pretty(render(decompose(MyBean("joe", 35))))
res0: String =
{
"name":"joe",
"age":35
}
But sometimes it is easier to use DSL syntax:
但有时使用 DSL 语法更容易：
scala> import net.liftweb.json.JsonDSL._
scala> val json = ("name" -> "joe") ~ ("age" -> 35)
scala> pretty(render(json))
res1: String =
{
"name":"joe",
"age":35
}
