Question

将应用程序从Swift2移到Swift3时，我遇到了一个错误，在尝试了几种不同的建议后，我一直无法修复。

lazy var address: AddressModel? = {
        [unowned self] in
        var dict = self.getpayloadDict()

        var model: AddressModel
        model = dict
        return model
        }()

model = dict throws无法分配类型为“ NSDictionary？”的值键入“ AddressModel”

AddressModel。。

class AddressModel: Deserializable {
    var City: String?
    var State: String?
    var PostalCode: String?    

    required init(data: [String: AnyObject]) {
        City = data["City"] as! String?
        State = data["State"] as! String?
        PostalCode = data["PostalCode"] as! String?
    }
}

任何帮助表示赞赏。

Answer 1

该错误也应该在Swift 2中发生。很明显：getpayloadDict()返回的字典与AddressModel不匹配。

您可以根据字典创建一个AddressModel实例

lazy var address: AddressModel? = { // this closure does not cause a retain cycle
    let dict = self.getpayloadDict()
    return AddressModel(data: dict)
}()

旁注：

as! String?（强制将可选内容展开为可选内容）是可怕的语法。使用常规的条件下调as? String。并且请遵循变量名以小写字母开头的命名约定。

Swift2到Swift3错误-无法分配'NSDictionary'类型的值？键入“ AddressModel”

1 个答案: