CODEWITHSUNDEEP
iosjsonstringswift
Fabian Boulegue
Fabian Boulegue

Reputation: 6608

Unwrapping Value after parsing JSON in Swift

I currently stuck at get my JSON Parse into a string, here is my Code

var SummName = "xxxxx"
        var APIKEY = "xxxx-xxx-43d5-8647-xxxx"



        let urlPath0 = "https://euw.api.pvp.net/api/lol/euw/v1.4/summoner/by-name/"+SummName+"?api_key="+APIKEY
        let url0 = NSURL(string: urlPath0)
        let session0 = NSURLSession.sharedSession()
        let task0 = session0.dataTaskWithURL(url0!, completionHandler: {data, response, error -> Void in
            if (error != nil) {
                println(error)
            } else {
                let summonorID_JSON = NSJSONSerialization.JSONObjectWithData(data, options: NSJSONReadingOptions.MutableContainers, error: nil) as NSDictionary
                var SummonerID = summonorID_JSON["smirknaitia"] as NSDictionary

                let SummID = SummonerID["id"]! as NSString
                println(SummID)
                let urlPath1 = "https://euw.api.pvp.net/api/lol/euw/v1.4/summoner/by-name/"+SummID+"?api_key="+APIKEY


            }
        })

        task0.resume()

It always fails by unwrapping the Value of SummonerID["ID"] into a string

Upvotes: 0

Views: 475

Answers (1)

radex
radex

Reputation: 6546

There are two ways let SummID = SummonerID["id"]! as NSString can fail:

  1. SummonerID dictionary doesn't have a value for key "id", so it fails when you try to force-unwrap
  2. SummonerID["id"]! is not of type NSString (or type that can convert to it, like String) so it fails when you try to force conversion.

We can't really tell which one it is exactly without seeing the input you have or exact error messages.

A side note: it's not in Swift style to start variable names with uppercase characters. It's best to use camelCase for variables, functions and methods; use UpperCamelCase for classes, structs, enums and enum cases.

Upvotes: 1

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