Why is the error "Unable to find encoder for type stored in a Dataset" when encoding JSON using case classes?

Milad Khajavi picture Milad Khajavi · Jan 11, 2016 · Viewed 21.1k times · Source

I've written spark job:

object SimpleApp {
  def main(args: Array[String]) {
    val conf = new SparkConf().setAppName("Simple Application").setMaster("local")
    val sc = new SparkContext(conf)
    val ctx = new org.apache.spark.sql.SQLContext(sc)
    import ctx.implicits._

    case class Person(age: Long, city: String, id: String, lname: String, name: String, sex: String)
    case class Person2(name: String, age: Long, city: String)

    val persons = ctx.read.json("/tmp/persons.json").as[Person]
    persons.printSchema()
  }
}

In IDE when I run the main function, 2 error occurs:

Error:(15, 67) Unable to find encoder for type stored in a Dataset.  Primitive types (Int, String, etc) and Product types (case classes) are supported by importing sqlContext.implicits._  Support for serializing other types will be added in future releases.
    val persons = ctx.read.json("/tmp/persons.json").as[Person]
                                                                  ^

Error:(15, 67) not enough arguments for method as: (implicit evidence$1: org.apache.spark.sql.Encoder[Person])org.apache.spark.sql.Dataset[Person].
Unspecified value parameter evidence$1.
    val persons = ctx.read.json("/tmp/persons.json").as[Person]
                                                                  ^

but in Spark Shell I can run this job without any error. what is the problem?

Answer

Developer picture Developer · Jan 11, 2016

The error message says that the Encoder is not able to take the Person case class.

Error:(15, 67) Unable to find encoder for type stored in a Dataset.  Primitive types (Int, String, etc) and Product types (case classes) are supported by importing sqlContext.implicits._  Support for serializing other types will be added in future releases.

Move the declaration of the case class outside the scope of SimpleApp.