CODEWITHSUNDEEP
jsonswiftnetwork-programmingnsdata
Andre
Andre

Reputation: 7668

Swift 3.0: Data to JSON [String : Any]

Evening, I'm trying to creating an APIClient, but I'm having a problem with a warning: APIClient.swift:53:81: Cast from 'Data' to unrelated type '[String : Any]' always fails

In this code I'm trying to convert Data into JSON as a dictionary [String : Any].

I guess the compiler can't know if this cast could or could not be possible so it throws the error, but I'm pretty sure it will work. So how can I avoid this warning or how can I write safer code?

do {
  let json = try JSONSerialization.data(
    withJSONObject: data!,
    options: []
  ) as? [String : Any]
  completion(json, HTTPResponse, nil)
} catch let error {
  completion(nil, HTTPResponse, error)
}

Upvotes: 41

Views: 79592

Answers (2)

Bilal Şimşek
Bilal Şimşek

Reputation: 5943

for Swift 5.*

with this function you can decode your api data to your codable struct, its also nice to catch to decoding errors detailed.

func decodeDataToObject<T: Codable>(data : Data?)->T?{
  if let data {
    do{
      return try JSONDecoder().decode(T.self, from: data)
    } catch let DecodingError.dataCorrupted(context) {
      print(context)
    } catch let DecodingError.keyNotFound(key, context) {
      print("Key '\(key)' not found:", context.debugDescription)
      print("codingPath:", context.codingPath)
    } catch let DecodingError.valueNotFound(value, context) {
      print("Value '\(value)' not found:", context.debugDescription)
      print("codingPath:", context.codingPath)
    } catch let DecodingError.typeMismatch(type, context)  {
      print("Type '\(type)' mismatch:", context.debugDescription)
      print("codingPath:", context.codingPath)
    } catch {
      print("error: ", error)
    }
  }
  return nil
}

usage; let user:User = decodeDataToObject(data: data)

Upvotes: 1

Andre
Andre

Reputation: 7668

The right method is:

do {  
    let json = try JSONSerialization.jsonObject(with: data!, options: []) as? [String : Any]
} catch {
    print("errorMsg")
}

Thanks to Eric Aya

Upvotes: 96

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