Syntax for passing “no options” to a parameter of type NSJSONWritingOptions

Question

In Objective-C I wrote code like the following:

NSError *error;
NSData *jsonData = [NSJSONSerialization dataWithJSONObject:someDictionary
                                                   options:0
                                                     error:&error];

I am trying to do the same thing in Swift. With some prompting from Xcode’s syntax checking I wrote it like this:

var jsonError: NSError
let jsonData = NSJSONSerialization.dataWithJSONObject(someDictionary, options: NSJSONWritingOptions(), error: &jsonError)

but this gives me an error: “Could not find an overload for init that accepts the supplied arguments.” I think the problem might be with the NSJSONWritingOptions() bit, and I’m guessing I just have the Swift syntax wrong. I tried replacing NSJSONWritingOptions() with NSJSONWritingOptions(0) and got the same error; I tried replacing it with nil (as suggested by this answer) but I got the error “Could not find an overload for __conversion that accepts the supplied arguments.”

How can I indicate that I want the default JSON writing options, whatever those might be?

score 4 · Accepted Answer · answered Jun 07 '14 at 16:32

4

The problem is not the NSJSONWritingOptions; the type of error you're passing should be NSError?, not NSError.

answered Jun 07 '14 at 16:32

Jesse Rusak

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\*sigh\* That fixed it, thanks. I hope they’ll be able to add a little more information to the error messages before the final release! – bdesham Jun 07 '14 at 16:36
Yeah, the errors are definitely a place that could use some work. – Jesse Rusak Jun 07 '14 at 16:39

Axel Guilmin · Answer 2 · 2018-11-28T09:48:17.613

3

For those who are looking for the Swift 4.2 equivalent, the options parameter is now optional:

var jsonData:Data?
do {
    jsonData = try JSONSerialization.data(withJSONObject: someDictionary)
} catch let parsingError {
    print(parsingError)
}

Or if you are 100% sure dataWithJSONObject will succeed:

let jsonData = try! JSONSerialization.data(withJSONObject:someDictionary)

edited Nov 28 '18 at 09:48

answered Oct 27 '15 at 08:36

Axel Guilmin

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In Swift 3+ *no options* is the default so the recommended way is to **omit** the parameter `JSONSerialization.data(withJSONObject:someDictionary)` – vadian Nov 28 '18 at 09:45

score 0 · Answer 3 · edited May 23 '17 at 12:33

0

Nicer way with swift will be:

let data = try! NSJSONSerialization.dataWithJSONObject(values, options: [])

to pass empty array...

source

edited May 23 '17 at 12:33

Community

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answered Dec 04 '15 at 18:06

Ben

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*Nicest* way is `let data = try! NSJSONSerialization.dataWithJSONObject(values)` – vadian Nov 28 '18 at 09:46

Syntax for passing “no options” to a parameter of type NSJSONWritingOptions

3 Answers3