Scala中List[obj]转成json对象

    技术2022-07-11  101

    使用如下方式转换的时候,对象的属性必须加上属性的标签,否则数据无法引入。 

    package flinkonkafka import com.alibaba.fastjson.JSON import scala.beans.BeanProperty import scala.collection.mutable.ListBuffer object test2{ def main(args: Array[String]) = { //第一种方式 import scala.collection.mutable.ListBuffer val result11: result1 = result1("Sister Ma", "Yoga") val result12: result1 = result1("Brother Bin", "work out") val result13: result1 = result1("Big chenyu","jokes") val result1s = ListBuffer(result11, result12, result13) val jstr: String = JSON.toJSONString(result1s.toArray,false) println(jstr) //第2种方式 // import org.json4s.{Formats,NoTypeHints} // import org.json4s.jackson.Serialization // import org.json4s.jackson.Serialization.write // // 需要添加隐式转换 // implicit val formats:AnyRef with Formats = Serialization.formats(NoTypeHints) // // 由scala对象转换为Json字符串 // val dstr = write(result1s) // println(dstr) } } case class result1(@BeanProperty code:String,@BeanProperty value:String) case class datar(@BeanProperty data:Object)

     

    Processed: 0.012, SQL: 9