scala - How to process multi line input records in Spark

Question

I have each record spread across multiple lines in the input file(Very huge file).

Ex:

Id:   2
ASIN: 0738700123
  title: Test tile for this product
  group: Book
  salesrank: 168501
  similar: 5  0738700811  1567184912  1567182813  0738700514  0738700915
  categories: 2
   |Books[283155]|Subjects[1000]|Religion & Spirituality[22]|Earth-Based Religions[12472]|Wicca[12484]
   |Books[283155]|Subjects[1000]|Religion & Spirituality[22]|Earth-Based Religions[12472]|Witchcraft[12486]
  reviews: total: 12  downloaded: 12  avg rating: 4.5
    2001-12-16  cutomer: A11NCO6YTE4BTJ  rating: 5  votes:   5  helpful:   4
    2002-1-7  cutomer:  A9CQ3PLRNIR83  rating: 4  votes:   5  helpful:   5

How to identify and process each multi line record in spark?

Answer 1

If the multi-line data has a defined record separator, you could use the hadoop support for multi-line records, providing the separator through a hadoop.Configuration object:

Something like this should do:

import org.apache.hadoop.conf.Configuration
import org.apache.hadoop.io.{LongWritable, Text}
import org.apache.hadoop.mapreduce.lib.input.TextInputFormat
val conf = new Configuration
conf.set("textinputformat.record.delimiter", "id:")
val dataset = sc.newAPIHadoopFile("/path/to/data", classOf[TextInputFormat], classOf[LongWritable], classOf[Text], conf)
val data = dataset.map(x=>x._2.toString)

This will provide you with an RDD[String] where each element corresponds to a record. Afterwards you need to parse each record following your application requirements.