CODEWITHSUNDEEP
iosswift
Mirko Brunner
Mirko Brunner

Reputation: 2215

Correct handling of NSJSONSerialization (try catch) in Swift (2.0)?

arowmy init works fine in Swift < 2 but in Swift 2 I get a error message from Xcode Call can throw, but it is not marked with 'try' and the error is not handled at let anyObj = NSJSONSerialization.JSONObjectWithData(data, options: NSJSONReadingOptions.MutableContainers) as! [String:AnyObject]. I think in my case I can´t use a try catch block because super is not initialized at this time. "Try" need a function that throws.

here is my function:

required init(coder aDecoder : NSCoder)
{
    self.name  = String(stringInterpolationSegment: aDecoder.decodeObjectForKey("name") as! String!)
    self.number = Int(aDecoder.decodeIntegerForKey("number"))
    self.img = String(stringInterpolationSegment: aDecoder.decodeObjectForKey("image") as! String!)
    self.fieldproperties = []

    var tmpArray = [String]()
    tmpArray = aDecoder.decodeObjectForKey("properties") as! [String]


    let c : Int = tmpArray.count
    for var i = 0; i < c; i++
    {
        let data : NSData = tmpArray[i].dataUsingEncoding(NSUTF8StringEncoding)!

         // Xcode(7) give me error: 'CAll can thorw, but it is not marked with 'try' and the error is not handled'
        let anyObj =  NSJSONSerialization.JSONObjectWithData(data, options: NSJSONReadingOptions.MutableContainers) as! [String:AnyObject]

        let label = anyObj["label"] as AnyObject! as! String
        let value = anyObj["value"] as AnyObject! as! Int
        let uprate = anyObj["uprate"] as AnyObject! as! Int
        let sufix = anyObj["sufix"] as AnyObject! as! String

        let props = Fieldpropertie(label: label, value: value, uprate: uprate, sufix: sufix)
        self.fieldproperties.append(props)
    }
}

Xcode mean that: let anyObj = try NSJSONSerialization.JSONObjectWithData(data, options: NSJSONReadingOptions.MutableContainers) as! [String:AnyObject]

but I have no idea to do here the right think according to this document https://developer.apple.com/library/content/documentation/Swift/Conceptual/Swift_Programming_Language/ErrorHandling.html

Upvotes: 30

Views: 32052

Answers (3)

Rob
Rob

Reputation: 437632

The jsonObject can throw errors, so put it within do block, use try, and catch any errors thrown. In Swift 3:

do {
    let anyObj = try JSONSerialization.jsonObject(with: data) as! [String: Any]

    let label = anyObj["label"] as! String
    let value = anyObj["value"] as! Int
    let uprate = anyObj["uprate"] as! Int
    let sufix = anyObj["sufix"] as! String

    let props = Fieldpropertie(label: label, value: value, uprate: uprate, sufix: sufix)

    // etc.
} catch {
    print("json error: \(error.localizedDescription)")
}

Or, in Swift 4, you can simplify your code by making your struct conform to Codable:

struct Fieldpropertie: Codable {
    let label: String
    let value: Int
    let uprate: Int
    let suffix: String
}

Then

do {
    let props = try JSONDecoder().decode(Fieldpropertie.self, from: data)
    // use props here; no manual parsing the properties is needed
} catch {
    print("json error: \(error.localizedDescription)")
}

For Swift 2, see previous revision of this answer.

Upvotes: 65

Gerd Castan
Gerd Castan

Reputation: 6849

JSONSerialization.JSONObject throws ErrorType and not NSError.

so the correct catch is 

do {
    let anyObj = try JSONSerialization.JSONObject(with: data, options: []) as! [String:AnyObject]
    // use anyObj here
} catch let error {
    print("json error: \(error)")
}

The type of error in catch let error is ErrorType

Upvotes: 5

trevorj
trevorj

Reputation: 2039

Don't know if it'll solve your problem, but isn't the method JSONObjectWithData:options:error:? I think you're missing the error parameter.

Upvotes: -3

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