Swift 4 JSONDecoder optional variable

Question

I have a Codable struct myObj:

public struct VIO: Codable {

    let id:Int?;
    ...
    var par1:Bool = false; //default to avoid error in parsing
    var par2:Bool = false;    
}

When I do receive JSON, I don't have par1 and par2 since these variables are optional. During parsing I get an error:keyNotFound(CodingKeys(stringValue: \"par1\", intValue: nil)

How to solve this?

vadian · Accepted Answer · 2018-04-07 19:20:04Z

3

If you have local variables you have to specify the CodingKeys

public struct VIO: Codable {

    private enum CodingKeys : String, CodingKey { case id }

    let id:Int?
    ...
    var par1:Bool = false
    var par2:Bool = false  

}

Edit:

If par1 and par2 should be also decoded optionally you have to write a custom initializer

  private enum CodingKeys : String, CodingKey { case id, par1, par2 }

   init(from decoder: Decoder) throws {
      let container = try decoder.container(keyedBy: CodingKeys.self)
      id = try container.decode(Int.self, forKey: .id)
      par1 = try container.decodeIfPresent(Bool.self, forKey: .par1)
      par2 = try container.decodeIfPresent(Bool.self, forKey: .par2)
  }

This is Swift: No trailing semicolons

edited Apr 7, 2018 at 19:20

answered Apr 7, 2018 at 16:35

vadian

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12 Comments

AlexP Over a year ago

so in my case I need to simply include private enum CodingKeys : String, CodingKey { case par1 = false case par2 = false } ?

Alexander Over a year ago

@AlexP No... what he showed is your case. The coding keys are usually inferred from the type's property names. Specifying the coding keys explicitly gives you a chance to omit par1, and par2, and just have id.

AlexP Over a year ago

this means in CodingKeys I shall put all variables that I am sure are present in JSON?

vadian Over a year ago

You need to specify only the CodingKeys you want to decode.

AlexP Over a year ago

I have a problem: now these values are never parsed even if present

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