How to parse JSON in Spark with fasterxml without

I got this far:

import com.fasterxml.jackson.module.scala.DefaultScalaModule
import com.fasterxml.jackson.module.scala.experimental.ScalaObjectMapper
import com.fasterxml.jackson.databind.ObjectMapper
import com.fasterxml.jackson.databind.DeserializationFeature

case class Person(name: String, lovesPandas: Boolean)

val mapper = new ObjectMapper()

val input = sc.textFile("files/pandainfo.json")
val result = input.flatMap(record => {
    try{
        Some(mapper.readValue(record, classOf[Person]))
    } catch {
        case e: Exception => None
    }
})
result.collect

but get Array() as a result (with no error). The file is https://github.com/databricks/learning-spark/blob/master/files/pandainfo.json How do I go on from here?

After consulting Spark: broadcasting jackson ObjectMapper I tried

import org.apache.spark._
import com.fasterxml.jackson.module.scala.DefaultScalaModule
import com.fasterxml.jackson.module.scala.experimental.ScalaObjectMapper
import com.fasterxml.jackson.databind.ObjectMapper
import com.fasterxml.jackson.databind.DeserializationFeature

case class Person(name: String, lovesPandas: Boolean)

val input = """{"name":"Sparky The Bear", "lovesPandas":true}"""
val result = input.flatMap(record => {
    try{
        val mapper = new ObjectMapper()
        mapper.registerModule(DefaultScalaModule)
        mapper.configure(DeserializationFeature.FAIL_ON_UNKNOWN_PROPERTIES, false)
        Some(mapper.readValue(record, classOf[Person]))
    } catch {
        case e: Exception => None
    }
})
result.collect

and got

Name: Compile Error
Message: <console>:34: error: overloaded method value readValue with alternatives:
  [T](x$1: Array[Byte], x$2: com.fasterxml.jackson.databind.JavaType)T <and>
  [T](x$1: Array[Byte], x$2: com.fasterxml.jackson.core.type.TypeReference[_])T <and>
  [T](x$1: Array[Byte], x$2: Class[T])T <and>

回答1:

I see that you tried the Learning Spark examples. Here the reference to the complete code https://github.com/holdenk/learning-spark-examples/blob/master/src/main/scala/com/oreilly/learningsparkexamples/scala/BasicParseJsonWithJackson.scala E.

回答2:

Instead of sc.textfile("path\to\json")you can try this(i write it in java cuz i don't know scala but the API is the same):

SQLContext sqlContext = new SQLContext(sc);
DataFrame dfFromJson = sqlContext.read().json("path\to\json\file.json");

spark will read your json file and convert it as a dataframe.

If your json file is nested, you could use

org.apache.spark.sql.functions.explode(e: Column): Column

for example, see my answer here

Hope this help you.