How to convert a JSON string to a dictionary?

Question

I want to make one function in my swift project that converts String to Dictionary json format but I got one error:

Cannot convert expression\'s type

Answer 1

Warning: this is a convenience method to convert a JSON string to a dictionary if, for some reason, you have to work from a JSON string. But if you have the JSON data available, you should instead work with the data, without using a string at all.

Swift 3

func convertToDictionary(text: String) -> [String: Any]? {
    if let data = text.data(using: .utf8) {
        do {
            return try JSONSerialization.jsonObject(with: data, options: []) as? [String: Any]
        } catch {
            print(error.localizedDescription)
        }
    }
    return nil
}

let str = "{\"name\":\"James\"}"

let dict = convertToDictionary(text: str)

Swift 2

func convertStringToDictionary(text: String) -> [String:AnyObject]? {
    if let data = text.dataUsingEncoding(NSUTF8StringEncoding) {
        do {
            return try NSJSONSerialization.JSONObjectWithData(data, options: []) as? [String:AnyObject]
        } catch let error as NSError {
            print(error)
        }
    }
    return nil
}

let str = "{\"name\":\"James\"}"

let result = convertStringToDictionary(str)

Original Swift 1 answer:

func convertStringToDictionary(text: String) -> [String:String]? {
    if let data = text.dataUsingEncoding(NSUTF8StringEncoding) {
        var error: NSError?
        let json = NSJSONSerialization.JSONObjectWithData(data, options: NSJSONReadingOptions.allZeros, error: &error) as? [String:String]
        if error != nil {
            println(error)
        }
        return json
    }
    return nil
}

let str = "{\"name\":\"James\"}"

let result = convertStringToDictionary(str) // ["name": "James"]

if let name = result?["name"] { // The `?` is here because our `convertStringToDictionary` function returns an Optional
    println(name) // "James"
}

In your version, you didn't pass the proper parameters to NSJSONSerialization and forgot to cast the result. Also, it's better to check for the possible error. Last note: this works only if your value is a String. If it could be another type, it would be better to declare the dictionary conversion like this:

let json = NSJSONSerialization.JSONObjectWithData(data, options: NSJSONReadingOptions.allZeros, error: &error) as? [String:AnyObject]

and of course you would also need to change the return type of the function:

func convertStringToDictionary(text: String) -> [String:AnyObject]? { ... }

Answer 2

With Swift 3, JSONSerialization has a method called json​Object(with:​options:​). json​Object(with:​options:​) has the following declaration:

class func jsonObject(with data: Data, options opt: JSONSerialization.ReadingOptions = []) throws -> Any

Returns a Foundation object from given JSON data.

When you use json​Object(with:​options:​), you have to deal with error handling (try, try? or try!) and type casting (from Any). Therefore, you can solve your problem with one of the following patterns.

---

#1. Using a method that throws and returns a non-optional type

import Foundation

func convertToDictionary(from text: String) throws -> [String: String] {
    guard let data = text.data(using: .utf8) else { return [:] }
    let anyResult: Any = try JSONSerialization.jsonObject(with: data, options: [])
    return anyResult as? [String: String] ?? [:]
}

Usage:

let string1 = "{\"City\":\"Paris\"}"
do {
    let dictionary = try convertToDictionary(from: string1)
    print(dictionary) // prints: ["City": "Paris"]
} catch {
    print(error)
}

let string2 = "{\"Quantity\":100}"
do {
    let dictionary = try convertToDictionary(from: string2)
    print(dictionary) // prints [:]
} catch {
    print(error)
}

let string3 = "{\"Object\"}"
do {
    let dictionary = try convertToDictionary(from: string3)
    print(dictionary)
} catch {
    print(error) // prints: Error Domain=NSCocoaErrorDomain Code=3840 "No value for key in object around character 9." UserInfo={NSDebugDescription=No value for key in object around character 9.}
}

---

#2. Using a method that throws and returns an optional type

import Foundation

func convertToDictionary(from text: String) throws -> [String: String]? {
    guard let data = text.data(using: .utf8) else { return [:] }
    let anyResult: Any = try JSONSerialization.jsonObject(with: data, options: [])
    return anyResult as? [String: String]
}

Usage:

let string1 = "{\"City\":\"Paris\"}"
do {
    let dictionary = try convertToDictionary(from: string1)
    print(String(describing: dictionary)) // prints: Optional(["City": "Paris"])
} catch {
    print(error)
}

let string2 = "{\"Quantity\":100}"
do {
    let dictionary = try convertToDictionary(from: string2)
    print(String(describing: dictionary)) // prints nil
} catch {
    print(error)
}

let string3 = "{\"Object\"}"
do {
    let dictionary = try convertToDictionary(from: string3)
    print(String(describing: dictionary))
} catch {
    print(error) // prints: Error Domain=NSCocoaErrorDomain Code=3840 "No value for key in object around character 9." UserInfo={NSDebugDescription=No value for key in object around character 9.}
}

---

#3. Using a method that does not throw and returns a non-optional type

import Foundation

func convertToDictionary(from text: String) -> [String: String] {
    guard let data = text.data(using: .utf8) else { return [:] }
    let anyResult: Any? = try? JSONSerialization.jsonObject(with: data, options: [])
    return anyResult as? [String: String] ?? [:]
}

Usage:

let string1 = "{\"City\":\"Paris\"}"
let dictionary1 = convertToDictionary(from: string1)
print(dictionary1) // prints: ["City": "Paris"]

let string2 = "{\"Quantity\":100}"
let dictionary2 = convertToDictionary(from: string2)
print(dictionary2) // prints: [:]

let string3 = "{\"Object\"}"
let dictionary3 = convertToDictionary(from: string3)
print(dictionary3) // prints: [:]

---

#4. Using a method that does not throw and returns an optional type

import Foundation

func convertToDictionary(from text: String) -> [String: String]? {
    guard let data = text.data(using: .utf8) else { return nil }
    let anyResult = try? JSONSerialization.jsonObject(with: data, options: [])
    return anyResult as? [String: String]
}

Usage:

let string1 = "{\"City\":\"Paris\"}"
let dictionary1 = convertToDictionary(from: string1)
print(String(describing: dictionary1)) // prints: Optional(["City": "Paris"])

let string2 = "{\"Quantity\":100}"
let dictionary2 = convertToDictionary(from: string2)
print(String(describing: dictionary2)) // prints: nil

let string3 = "{\"Object\"}"
let dictionary3 = convertToDictionary(from: string3)
print(String(describing: dictionary3)) // prints: nil

Answer 3

Swift 3:

if let data = text.data(using: String.Encoding.utf8) {
    do {
        let json = try JSONSerialization.jsonObject(with: data, options: .mutableContainers) as? [String:Any]
        print(json)
    } catch {
        print("Something went wrong")
    }
}

Answer 4

I've updated Eric D's answer for Swift 5:

 func convertStringToDictionary(text: String) -> [String:AnyObject]? {
    if let data = text.data(using: .utf8) {
        do {
            let json = try JSONSerialization.jsonObject(with: data, options: .mutableContainers) as? [String:AnyObject]
            return json
        } catch {
            print("Something went wrong")
        }
    }
    return nil
}

Answer 5

Swift 5

extension String {
    func convertToDictionary() -> [String: Any]? {
        if let data = data(using: .utf8) {
            return try? JSONSerialization.jsonObject(with: data, options: []) as? [String: Any]
        }
        return nil
    }
}

Answer 6

Swift 4

extension String {
    func convertToDictionary() -> [String: Any]? {
        if let data = self.data(using: .utf8) {
            do {
                return try JSONSerialization.jsonObject(with: data, options: []) as? [String: Any]
            } catch {
                print(error.localizedDescription)
            }
        }
        return nil
    }
}