Scala: parse JSON file into List[DBObject]

Question

1.Input is JSON file that contains multiple records. Example:

[ {"user": "user1", "page": 1, "field": "some"}, {"user": "user2", "page": 2, "field": "some2"}, ... ]

2.I need to load each record from the file as a Document to MongoDB collection. Using casbah for interacting with mongo, inserting data may look like:

  def saveCollection(inputListOfDbObjects: List[DBObject]) = {
    val xs = inputListOfDbObjects
    xs foreach (obj => {
    Collection.save(obj)
  })

Question: What is the correct way (using scala) to parse JSON to get data as List[DBObject] at output?

Any help is appreciated.

travisdazell · Answer 1 · 2016-07-20T19:19:27.150

You could use the parser combinator library in Scala.

Here's some code I found that does this for JSON: http://booksites.artima.com/programming_in_scala_2ed/examples/html/ch33.html#sec4

Step 1. Create a class named JSON that contains your parser rules:

import scala.util.parsing.combinator._

class JSON extends JavaTokenParsers {   

  def value : Parser[Any] = obj | arr | 
                        stringLiteral | 
                        floatingPointNumber | 
                        "null" | "true" | "false"

  def obj   : Parser[Any] = "{"~repsep(member, ",")~"}"

  def arr   : Parser[Any] = "["~repsep(value, ",")~"]"

  def member: Parser[Any] = stringLiteral~":"~value
}

Step 2. In your main function, read in your JSON file, passing the contents of the file to your parser.

import java.io.FileReader

object ParseJSON extends JSON {
    def main(args: Array[String]) {
        val reader = new FileReader(args(0))
        println(parseAll(value, reader))
    }
}

It's a good idea to put the code into your answer. Links break all the time, which means your answer would no longer work. — Hyperbole, Jul 20 '16 at 19:12

Scala: parse JSON file into List[DBObject]

1 Answers1