使用lift-json将Scala对象序列化为JSon字符串 [英] Serializing a scala object into a JSon String using lift-json

问题描述

我想知道，请您告诉我如何使用lift-json将简单的bean类序列化为json字符串(我使用的是v2.0-M1). 我尝试过:

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:

val r = JsonDSL.pretty(JsonAST.render(myBean))

我得到了

[error]  found   : MyBean
[error]  required: net.liftweb.json.JsonAST.JValue

谢谢， -A

推荐答案

您可以将案例类分解"为JSON，然后进行呈现.示例:

You can "decompose" a case class into JSON and then render it. Example:

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
}

但是有时候使用DSL语法会更容易:

But sometimes it is easier to use DSL syntax:

scala> import net.liftweb.json.JsonDSL._
scala> val json = ("name" -> "joe") ~ ("age" -> 35)
scala> pretty(render(json))
res1: String = 
{
  "name":"joe",
  "age":35
}

这篇关于使用lift-json将Scala对象序列化为JSon字符串的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！