When you performed the migration Xcode asked about @objc inference and you probably selected the new type instead of Swift3. Possible solutions: Use @objc Use @objc on each method, as needed instead of the whole class. Use @objcMembers You could just use @objcMembers on the class. Applying @objcMembers attribute to a class implicitly adds the … Read more
The syntax is simply: // to run something in 0.1 seconds DispatchQueue.main.asyncAfter(deadline: .now() + 0.1) { // your code here } Note, the above syntax of adding seconds as a Double seems to be a source of confusion (esp since we were accustomed to adding nsec). That “add seconds as Double” syntax works because deadline … Read more
I believe in the case of inheritance you must implement Coding yourself. That is, you must specify CodingKeys and implement init(from:) and encode(to:) in both superclass and subclass. Per the WWDC video (around 49:28, pictured below), you must call super with the super encoder/decoder. required init(from decoder: Decoder) throws { // Get our container for … Read more
I got rid of this warning by changing the “Swift 3 @objc Inference” build setting of my targets to “Default”. From this article: Before Swift 4, the compiler made some Swift declarations automatically available to Objective-C. For example, if one subclassed from NSObject, the compiler created Objective-C entry points for all methods in such classes. … Read more
With some inspiration from this gist I found, I wrote some extensions for UnkeyedDecodingContainer and KeyedDecodingContainer. You can find a link to my gist here. By using this code you can now decode any Array<Any> or Dictionary<String, Any> with the familiar syntax: let dictionary: [String: Any] = try container.decode([String: Any].self, forKey: key) or let array: … Read more
You should leave one side empty, hence the name “partial range”. let newStr = str[..<index] The same stands for partial range from operators, just leave the other side empty: let newStr = str[index…] Keep in mind that these range operators return a Substring. If you want to convert it to a string, use String‘s initialization … Read more
Another approach is to create an intermediate model that closely matches the JSON (with the help of a tool like quicktype.io), let Swift generate the methods to decode it, and then pick off the pieces that you want in your final data model: // snake_case to match the JSON and hence no need to write … Read more
struct GeneralProduct: Codable { var price: Double? var id: String? var name: String? private enum CodingKeys: String, CodingKey { case price = “p”, id = “i”, name = “n” } init(price: Double? = nil, id: String? = nil, name: String? = nil) { self.price = price self.id = id self.name = name } init(from decoder: … Read more
The key is in how you define the CodingKeys property. While it’s most commonly an enum it can be anything that conforms to the CodingKey protocol. And to make dynamic keys, you can call a static function: struct Category: Decodable { struct Detail: Decodable { var category: String var trailerPrice: String var isFavorite: Bool? var … Read more
You can use the Codable interface with CoreData objects to encode and decode data, however it’s not as automatic as when used with plain old swift objects. Here’s how you can implement JSON Decoding directly with Core Data objects: First, you make your object implement Codable. This interface must be defined on the object, and … Read more