Type "Any" has no subscript members while parsing JSON

Question

func parseResponseData(response: AnyObject) {
    if let feedData = response["feed"] as? [String: Any]{
        let entryArray = feedData["entry"] as Any
        if let entry = entryArray as? [Any]{
            for object in entry{
                print(object)
                let appName = object["category"] as [String: Any]  // Error: Type "Any" has no subscript members.
            }   
        } 
    }
}

I am not able to parse the response because of this error, any clue how to resolve this.

JSON response

Possible duplicate of [Correctly Parsing JSON in Swift 3](http://stackoverflow.com/questions/39423367/correctly-parsing-json-in-swift-3) — Larme, May 11 '17 at 09:47

score 0 · Answer 1 · answered May 11 '17 at 07:53

0

try this, entryArray is an array of String : Any

func parseResponseData(response: AnyObject) {
    if let feedData = response["feed"] as? [String: Any]{
        let entryArray = feedData["entry"] as Any
        if let entry = entryArray as? [[String: Any]] {
            for object in entry {
                print(object)
                let appName = object["category"] as? [String: Any]
            }
        }
    }
}

answered May 11 '17 at 07:53

JuicyFruit

2,638
2
18
35

It Worked! Thanks @JuicyFruit for the type casting. That was really helpful. – Saikat Dey May 11 '17 at 13:08

score 0 · Answer 2 · answered May 11 '17 at 08:31

You have to tell the compiler the actual types of all subscripted collection types, Any is not sufficient.

It's helpful to use a type alias for the JSON dictionary

typealias JSONDictionary = [String:Any]

func parseResponseData(response: Any) {
    if let feedData = response["feed"] as? JSONDictionary,
     let entryArray = feedData["entry"] as? [JSONDictionary] {
        for object in entryArray {
            print(object)
            if let category = object["category"] as? JSONDictionary {
                print(category)
            }
        } 
    }
}

Type "Any" has no subscript members while parsing JSON

2 Answers2