CODEWITHSUNDEEP
iosjsonswiftswift4swifty-json
Qedrix
Qedrix

Reputation: 473

Invalid type in JSON write (__SwiftValue)

I am trying to pass a Parameters [String:Any] value to JSONSerialization.data and it's throwing error every time.

I know the values of the [String:Any] dictionary is Swifty.JSON objects. But I am not able to convert them to NSDictionary object.

I populate the params from another dictionary like so:

var params = [String:Any]()
for (key, value) in self.myDictionary[self.selectedEntry] as! JSON {
    print("\(key) - \(value.description)")
    params[key]=value     
}

And this is whats inside the params object after a print(params).

["searchOptions": {
  "omit_saved_records" : false
}, "customFilters": {
  "is_next_sellers_search" : "Y"
}, "use_code_std": [
  "RSFR",
  "RCON"
]]

I am passing the params to this function:

let json = try JSONSerialization.data(withJSONObject: params, options: .prettyPrinted)

This is where the error occur.

I am expecting this to simply work but I get this error instead:

Terminating app due to uncaught exception 'NSInvalidArgumentException', reason: 'Invalid type in JSON write (__SwiftValue)'

What am I doing wrong here?

Upvotes: 2

Views: 10737

Answers (1)

vadian
vadian

Reputation: 285079

The JSON type of SwiftyJSON is a custom type and cannot be serialized.

Just get the dictionaryObject from the JSON object

if let params = self.myDictionary[self.selectedEntry].dictionaryObject {

And don't prettyPrint. The server doesn't care

let json = try JSONSerialization.data(withJSONObject: params)

Note: You are encouraged to drop SwiftyJSON in favor of Codable.

Upvotes: 5

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