How to convert array of dictionary to JSON?

Question

I have an array of dictionaries that I\'d like to convert to JSON. My object is of type [[String: AnyObject]] and would like to end up with a sample like this:<

Answer 1

Swift 4:

extension Collection where Iterator.Element == [String:AnyObject] {
    func toJSONString(options: JSONSerialization.WritingOptions = .prettyPrinted) -> String {
        if let arr = self as? [[String:AnyObject]],
            let dat = try? JSONSerialization.data(withJSONObject: arr, options: options),
            let str = String(data: dat, encoding: String.Encoding.utf8) {
            return str
        }
        return "[]"
    }
}

Usage:

let arrayOfDictionaries = [
  { "abc": 123, "def": "ggg", "xyz": true },
  { "abc": 456, "def": "hhh", "xyz": false },
  { "abc": 789, "def": "jjj", "xyz": true }
]

print(arrayOfDictionaries.toJSONString())

Answer 2

A simple way to achieve that is to just extend CollectionType.

Use optional binding and downcasting, then serialize to data, then convert to string.

extension CollectionType where Generator.Element == [String:AnyObject] {
    func toJSONString(options: NSJSONWritingOptions = .PrettyPrinted) -> String {
        if let arr = self as? [[String:AnyObject]],
            let dat = try? NSJSONSerialization.dataWithJSONObject(arr, options: options),
            let str = String(data: dat, encoding: NSUTF8StringEncoding) {
            return str
        }
        return "[]"
    }
}

let arrayOfDictionaries: [[String:AnyObject]] = [
    ["abc":123, "def": "ggg", "xyz": true],
    ["abc":456, "def": "hhh", "xyz": false]
]

print(arrayOfDictionaries.toJSONString())

Output:

[
  {
    "abc" : 123,
    "def" : "ggg",
    "xyz" : true
  },
  {
    "abc" : 456,
    "def" : "hhh",
    "xyz" : false
  }
]

Answer 3

I think you're out of luck trying to extend Array, because same-type requirements are not possible when extending an entity.

What you can make is to make use of free functions:

func jsonify(array: [[String:AnyObject]]) { 
...
}

This way you'll keep the type safety you want to, with the expense of moving the function location.