How to format a JSON String in Swift? - json

Question

I have a JSON String that follows a format like { name: "John" } and not { "name" : "John"} and that is resulting in a nil whenever I try to access the name key because:
Error Domain=NSCocoaErrorDomain Code=3840 "No string key for value in object around character 1."
I am looking for a function that fixes/parses/formats this JSON file into something readable? How do sites like JSON Format do it?

Answer 1

Funny enough, { name: "John" } makes a valid JSON object in Javascript. So your problem now becomes finding a Javascript intepreter for Swift!
There's one built-in to recent versions of Mac OS X and iOS: WKWebView. It's a web rendering engine with a Javascript parser. Link your target with WebKit and try this:
import WebKit
class MyJSONParser {
private static let webView = WKWebView()
class func parse(jsonString: String, completionHandler: (AnyObject?, NSError?) -> Void) {
self.webView.evaluateJavaScript(jsonString, completionHandler: completionHandler)
}
}
Usage:
let str = "{ firstName: 'John', lastName: 'Smith' }"
// You must assign the JSON string to a variable or the Javascript
// will return void. Note that this runs asynchronously
MyJSONParser.parse("tmp = \(str)") { result, error in
guard error == nil else {
print(error)
return
}
if let dict = result as? [String: String] {
print(dict)
} else {
print("Can't convert to Dictionary")
}
}
Swift 3
import WebKit
class MyJSONParser {
private static let webView = WKWebView()
class func parse(jsonString: String, completionHandler: #escaping (Any?, Error?) -> Void) {
self.webView.evaluateJavaScript(jsonString, completionHandler: completionHandler)
}
}
let str = "{ firstName: 'John', lastName: 'Smith' }"
// You must assign the JSON string to a variable or the Javascript
// will return void. Note that this runs asynchronously
MyJSONParser.parse(jsonString: "tmp = \(str)") { result, error in
guard error == nil else {
print(error!)
return
}
if let dict = result as? [String: String] {
print(dict)
} else {
print("Can't convert to Dictionary")
}
}