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Scala generics usage for methods + json4s parsing

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0

I am not sure if this is achievable and I have a very basic understanding of how generics work in scala. But I was wondering if this is possible.
Say I have a method:

case class Person(id:String,name:String)
case class Student(id:String,name:String, class:String)

 def convertToJson[A](fileName:String):
 //read file
 parse[A]
 

Is it possible to write this generic method which would parse the json based on the type of class I send when I call the convertToJson method?
Something like:

convertToJson[Student](fileName)
convertToJson[Person](fileName)

BTW the above code gives me a :

No Manifest available for A. error.

Using json4s for parsing.
Any help is appreciated.

scala generics json4s

share|improve this question

asked Mar 26 at 16:33

AchilleusAchilleus

86955 silver badges2020 bronze badges

add a comment | 

0

I am not sure if this is achievable and I have a very basic understanding of how generics work in scala. But I was wondering if this is possible.
Say I have a method:

case class Person(id:String,name:String)
case class Student(id:String,name:String, class:String)

 def convertToJson[A](fileName:String):
 //read file
 parse[A]
 

Is it possible to write this generic method which would parse the json based on the type of class I send when I call the convertToJson method?
Something like:

convertToJson[Student](fileName)
convertToJson[Person](fileName)

BTW the above code gives me a :

No Manifest available for A. error.

Using json4s for parsing.
Any help is appreciated.

scala generics json4s

share|improve this question

asked Mar 26 at 16:33

AchilleusAchilleus

86955 silver badges2020 bronze badges

add a comment | 

I am not sure if this is achievable and I have a very basic understanding of how generics work in scala. But I was wondering if this is possible.
Say I have a method:

case class Person(id:String,name:String)
case class Student(id:String,name:String, class:String)

 def convertToJson[A](fileName:String):
 //read file
 parse[A]
 

Is it possible to write this generic method which would parse the json based on the type of class I send when I call the convertToJson method?
Something like:

convertToJson[Student](fileName)
convertToJson[Person](fileName)

BTW the above code gives me a :

No Manifest available for A. error.

Using json4s for parsing.
Any help is appreciated.

scala generics json4s

share|improve this question

asked Mar 26 at 16:33

AchilleusAchilleus

86955 silver badges2020 bronze badges

case class Person(id:String,name:String)
case class Student(id:String,name:String, class:String)

 def convertToJson[A](fileName:String):
 //read file
 parse[A]
 

convertToJson[Student](fileName)
convertToJson[Person](fileName)

share|improve this question

asked Mar 26 at 16:33

AchilleusAchilleus

86955 silver badges2020 bronze badges

share|improve this question

asked Mar 26 at 16:33

AchilleusAchilleus

86955 silver badges2020 bronze badges

asked Mar 26 at 16:33

AchilleusAchilleus

86955 silver badges2020 bronze badges

asked Mar 26 at 16:33

AchilleusAchilleus

86955 silver badges2020 bronze badges

1 Answer
1

active

oldest

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3

This will convert a JSON string to a case class

import org.json4s._
import org.json4s.jackson.JsonMethods._

def convertToJson[T](json: String)(implicit fmt: Formats = DefaultFormats, mf: Manifest[T]): T =
 Extraction.extract(parse(json))

Once this is defined you can parse appropriate strings to the required type:

case class Person(id: String, name: String)
case class Student(id: String, name: String, `class`: String)

val person = convertToJson[Person](""""name":"Jane","id":45""")
val student = convertToJson[Student](""""name":"John","id":63, "class": "101"""")

Note that this will ignore JSON data that does not match fields in the case class. If a field is optional in the JSON, make it an Option in the case class and you will get None if the field is not there.

share|improve this answer

answered Mar 26 at 18:53

TimTim

9,20511 gold badge88 silver badges2020 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  That worked! Thanks. On a side note is there a way to deal with nulling out case when we have bad data? Say throw an exception or something?
  
  – Achilleus
  Mar 26 at 19:52
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @Achilleus If you have bad data the extract will throw an exception. You can let the caller handle that, or wrap the extract call in a Try and return a Try[T], which is perhaps more standard in Scala.
  
  – Tim
  Mar 26 at 23:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Actually, it won't throw an exception in the situation when we are dealing with nested json and an object in it say has a field spelled wrong. It will just null that object and move on. I was wondering if we can somehow specify it to fail or throw an exception instead of nulling it out.
  
  – Achilleus
  Mar 26 at 23:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Can you give an example of this? If I rename one of the fields from the JSON string I get MappingException: No usable value for <fieldname>.
  
  – Tim
  Mar 26 at 23:50
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Ah actually, it throws the exception for case classes. But I am using an Avro specific record case class. Its a case class that extends org.apache.avro.specific.SpecificRecordBase. But somehow it serializes to null in this case.
  
  – Achilleus
  Mar 27 at 16:05
  

  
  
  
  

 | 
show 1 more comment

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3

This will convert a JSON string to a case class

import org.json4s._
import org.json4s.jackson.JsonMethods._

def convertToJson[T](json: String)(implicit fmt: Formats = DefaultFormats, mf: Manifest[T]): T =
 Extraction.extract(parse(json))

Once this is defined you can parse appropriate strings to the required type:

case class Person(id: String, name: String)
case class Student(id: String, name: String, `class`: String)

val person = convertToJson[Person](""""name":"Jane","id":45""")
val student = convertToJson[Student](""""name":"John","id":63, "class": "101"""")

Note that this will ignore JSON data that does not match fields in the case class. If a field is optional in the JSON, make it an Option in the case class and you will get None if the field is not there.

share|improve this answer

answered Mar 26 at 18:53

TimTim

9,20511 gold badge88 silver badges2020 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  That worked! Thanks. On a side note is there a way to deal with nulling out case when we have bad data? Say throw an exception or something?
  
  – Achilleus
  Mar 26 at 19:52
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @Achilleus If you have bad data the extract will throw an exception. You can let the caller handle that, or wrap the extract call in a Try and return a Try[T], which is perhaps more standard in Scala.
  
  – Tim
  Mar 26 at 23:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Actually, it won't throw an exception in the situation when we are dealing with nested json and an object in it say has a field spelled wrong. It will just null that object and move on. I was wondering if we can somehow specify it to fail or throw an exception instead of nulling it out.
  
  – Achilleus
  Mar 26 at 23:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Can you give an example of this? If I rename one of the fields from the JSON string I get MappingException: No usable value for <fieldname>.
  
  – Tim
  Mar 26 at 23:50
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Ah actually, it throws the exception for case classes. But I am using an Avro specific record case class. Its a case class that extends org.apache.avro.specific.SpecificRecordBase. But somehow it serializes to null in this case.
  
  – Achilleus
  Mar 27 at 16:05
  

  
  
  
  

 | 
show 1 more comment

3

This will convert a JSON string to a case class

import org.json4s._
import org.json4s.jackson.JsonMethods._

def convertToJson[T](json: String)(implicit fmt: Formats = DefaultFormats, mf: Manifest[T]): T =
 Extraction.extract(parse(json))

Once this is defined you can parse appropriate strings to the required type:

case class Person(id: String, name: String)
case class Student(id: String, name: String, `class`: String)

val person = convertToJson[Person](""""name":"Jane","id":45""")
val student = convertToJson[Student](""""name":"John","id":63, "class": "101"""")

Note that this will ignore JSON data that does not match fields in the case class. If a field is optional in the JSON, make it an Option in the case class and you will get None if the field is not there.

share|improve this answer

answered Mar 26 at 18:53

TimTim

9,20511 gold badge88 silver badges2020 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  That worked! Thanks. On a side note is there a way to deal with nulling out case when we have bad data? Say throw an exception or something?
  
  – Achilleus
  Mar 26 at 19:52
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @Achilleus If you have bad data the extract will throw an exception. You can let the caller handle that, or wrap the extract call in a Try and return a Try[T], which is perhaps more standard in Scala.
  
  – Tim
  Mar 26 at 23:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Actually, it won't throw an exception in the situation when we are dealing with nested json and an object in it say has a field spelled wrong. It will just null that object and move on. I was wondering if we can somehow specify it to fail or throw an exception instead of nulling it out.
  
  – Achilleus
  Mar 26 at 23:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Can you give an example of this? If I rename one of the fields from the JSON string I get MappingException: No usable value for <fieldname>.
  
  – Tim
  Mar 26 at 23:50
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Ah actually, it throws the exception for case classes. But I am using an Avro specific record case class. Its a case class that extends org.apache.avro.specific.SpecificRecordBase. But somehow it serializes to null in this case.
  
  – Achilleus
  Mar 27 at 16:05
  

  
  
  
  

 | 
show 1 more comment

This will convert a JSON string to a case class

import org.json4s._
import org.json4s.jackson.JsonMethods._

def convertToJson[T](json: String)(implicit fmt: Formats = DefaultFormats, mf: Manifest[T]): T =
 Extraction.extract(parse(json))

Once this is defined you can parse appropriate strings to the required type:

case class Person(id: String, name: String)
case class Student(id: String, name: String, `class`: String)

val person = convertToJson[Person](""""name":"Jane","id":45""")
val student = convertToJson[Student](""""name":"John","id":63, "class": "101"""")

Note that this will ignore JSON data that does not match fields in the case class. If a field is optional in the JSON, make it an Option in the case class and you will get None if the field is not there.

share|improve this answer

answered Mar 26 at 18:53

TimTim

9,20511 gold badge88 silver badges2020 bronze badges

import org.json4s._
import org.json4s.jackson.JsonMethods._

def convertToJson[T](json: String)(implicit fmt: Formats = DefaultFormats, mf: Manifest[T]): T =
 Extraction.extract(parse(json))

case class Person(id: String, name: String)
case class Student(id: String, name: String, `class`: String)

val person = convertToJson[Person](""""name":"Jane","id":45""")
val student = convertToJson[Student](""""name":"John","id":63, "class": "101"""")

share|improve this answer

answered Mar 26 at 18:53

TimTim

9,20511 gold badge88 silver badges2020 bronze badges

answered Mar 26 at 18:53

TimTim

9,20511 gold badge88 silver badges2020 bronze badges

answered Mar 26 at 18:53

TimTim

9,20511 gold badge88 silver badges2020 bronze badges