Question

以下代码抛出一条消息，指出“条件绑定的初始化程序必须具有可选类型，而不是'AnyObject'”

    func parseData2(){
        var data:NSData?


        if let data2 = data  {
            do {
                let details = try NSJSONSerialization.JSONObjectWithData(data2, options: .AllowFragments)

                if let actualDetails = details where actualDetails.isKindOfClass(NSDictionary) {
                    print("Parse Data")
                }

            }catch {
                print("Error \(error)")
            }
        }

    }

要解决上述错误，我使用了以下代码。

    func parseData2(){
        var data:NSData?


        if let data2 = data  {
            do {
                let details:AnyObject = try NSJSONSerialization.JSONObjectWithData(data2, options: .AllowFragments)

                if let actualDetails:AnyObject = details where actualDetails.isKindOfClass(NSDictionary) {
                    print("Parse Data")
                }

            }catch {
                print("Error \(error)")
            }
        }

    }

有没有比上面更好的方法或我的代码可能会崩溃？

我想添加一个代码，考虑零检查，类型检查，然后输入检查。 Swift背后的原因提供了很大的灵活性，但很难解决问题。假设我有一个字典， cityDetails ，我试图获取self.cityZipCode和self.cityIdentifier的数据，这些数据是可选的，定义为var cityZipCode：Int？和var cityIdentifier：Int？

if let cityBasic = cityDetails["basicDetails"] where 
cityBasic!.isKindOfClass(NSDictionary) {

self.cityZipCode = (cityBasic as! NSDictionary)["zip"].integerValue ?? 0

self.cityIdentifier =  (cityBasic as! NSDictionary)["cityId"].integerValue ?? 0

}

Answer 1

无需从try打开结果。它不是可选的。您需要将结果从try投射到NSDictionary。使用as?向下转发。

最佳做法：完全访问返回的错误，以便进行良好的错误处理

func parseData2(){
    var data:NSData?

    if let data2 = data  {
        do {
            let details = try NSJSONSerialization.JSONObjectWithData(data2, options: .AllowFragments)

            if let detailsDict = details as? NSDictionary {
                print("Parse Data")
            } else if let detailsArray = details as? NSArray {
                print("array")
            }

        } catch {
            print("Error \(error)")
        }
    }
}

快速而肮脏：错误处理不适合我！

func parseData2(){
    var data:NSData?

    if let data2 = data  {

        let details = try? NSJSONSerialization.JSONObjectWithData(data2, options: .AllowFragments)

        if let detailsDict = details as? NSDictionary {
            print("Parse Data")
        } else {
            print("details might be nil, or not an NSDictionary")
        }
    }
}

糟糕的屁股模式：崩溃是功能

func parseData2(){
    var data:NSData?

    if let data2 = data  {

        let details = try! NSJSONSerialization.JSONObjectWithData(data2, options: .AllowFragments) as! NSDictionary

    }
}

多个unwraps的一些额外信息：将以下代码放在游乐场中。

struct SomeStruct {
    var anOptional : Int?
    init() {
    }
}

func unwrapWithIfLet() {

    if let unWrappedStruct = myStruct, let unWrappedSomething = unWrappedStruct.anOptional {
        print("multiple optional bindings succeeded")
        // both unWrappedStruct and unWrappedSomething are available here
    } else {
        print("something is nil")
    }
}

func unwrapWithGuard() {

    guard let unWrappedStruct = myStruct, let unWrappedSomething = unWrappedStruct.anOptional else {
        print("something is nil")
        return
    }
    print("multiple optional bindings succeeded")
    // both unWrappedStruct and unWrappedSomething are available here
}


var myStruct : SomeStruct?

//unwrapWithGuard()
//unwrapWithIfLet()

myStruct = SomeStruct()
myStruct!.anOptional = 1

unwrapWithGuard()
unwrapWithIfLet()

Answer 2

您正在寻找as?，它会尝试将左侧的内容转换为右侧的内容，如果转换不可能则返回nil：

let details = try NSJSONSerialization.JSONObjectWithData(data2, options: .AllowFragments)

if let actualDetails = details as? NSDictionary {
    print("Parse Data")
}

您很少需要在Swift中使用isKindOfClass。如果您发现自己使用它，请询问原因，并考虑as或as?是否有效。

条件绑定的初始化程序必须具有可选类型，而不是'AnyObject - Approach

2 个答案: