Swift: testing against optional value in switch case

Question

In Swift, how can I write a case in a switch statement that tests the value being switched against the contents of an optional, skipping over the case if the option

Answer 1

In Swift 4 you can use Optional : ExpressibleByNilLiteral of Apple to wrappe optional

https://developer.apple.com/documentation/swift/optional

Example

enum MyEnum {
    case normal
    case cool
}

some

let myOptional: MyEnum? = MyEnum.normal

switch smyOptional {
    case .some(.normal): 
    // Found .normal enum
    break

    case .none: 
    break

    default:
    break
}

none

let myOptional: MyEnum? = nil

switch smyOptional {
    case .some(.normal): 
    break

    case .none: 
    // Found nil
    break

    default:
    break
}

default

let myOptional: MyEnum? = MyEnum.cool

switch smyOptional {
    case .some(.normal): 
    break

    case .none: 
    break

    default:
    // Found .Cool enum
    break
}

Enum with value

enum MyEnum {
    case normal(myValue: String)
    case cool
}

some value

let myOptional: MyEnum? = MyEnum.normal("BlaBla")

switch smyOptional {
case .some(.normal(let myValue)) where myValue == "BlaBla":
    // Here because where find in my myValue "BlaBla"
    break

// Example for get value
case .some(.normal(let myValue)):
    break

// Example for just know if is normal case enum
case .some(.normal):
    break

case .none:
    break

default:

    break
}