How to create a embedded JSON Object and return to String in Swift?

Question

My code.swift is below :

public class Test {

    var testA: String = ""                
    var testB: String = ""                 

    let T_testA : String = "testA"
    let T_testB : String = "testB"

    init(testA: String, testB: String) {
        self.testA = testA
        self.testB = testB
    }

    func toString() -> String? {
        let jsonDic : [String: AnyObject] = [
            T_testA: testA,
            T_testB: testB,
        ]
        do {
            let jsonObject = try NSJSONSerialization.dataWithJSONObject( jsonDic, options: NSJSONWritingOptions.PrettyPrinted)

            return String(data: jsonObject, encoding: NSUTF8StringEncoding)
        } catch {
            return nil
        }
    }
}

In my Tests.swift

func testPerformanceExample() {

    let result1 = Test(testA: {"major": 1, "minor": 2}, testB: "http://google.com")

    let result2 = result2.toString()!
    print("result2=\n\(result2)")
}

I read similar case Here , It seems a different problem!

Is that possible to create the embedded json object together and return it as a String ?

In the class the property testA is declared as String. How can testA in Tests be a dictionary? — vadian
– vadian, Commented Apr 27, 2016 at 10:05
You should not create JSON String from a Swift String. Instead, make a dictionary/array then convert it to JSON data then convert it back to JSON String. There's examples in my answers - and of course many others in SO. — Eric Aya
– Eric Aya, Commented Apr 27, 2016 at 10:09

Eric Aya · Accepted Answer · 2016-04-27 10:18:16Z

1

If you want to get something from your existing code, use it properly:

let result1 = Test(testA: "{\"major\": 1, \"minor\": 2}", testB: "http://google.com")
let result2 = result1.toString()!
print("result2=\n\(result2)")

But I actually suggest you study this other way instead: make a Swift dictionary, convert this dictionary to JSON data, then convert this data to a String. Then you will get a correct JSON String without risking writing erroneous Strings.

answered Apr 27, 2016 at 10:18

Eric Aya

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1 Comment

HungCLo Over a year ago

Thanks for your help, I think you suggestion is really helpful.

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