swift - 我如何将字典作为函数参数传递？

Question

我正在尝试将字典作为函数参数传递。我有以下功能

func makeAndAddVisitorRecord2(visitorDict: Dictionary) -> ABRecordRef <AnyObject, AnyObject> {
    let visitorRecord: ABRecordRef = ABPersonCreate().takeRetainedValue()
    ABRecordSetValue(visitorRecord, kABPersonFirstNameProperty, visitorDict[1], nil)
    ABRecordSetValue(visitorRecord, kABPersonLastNameProperty, visitorDict[2], nil)
    //ABRecordSetValue(visitorRecord, kABPersonEmailProperty, visitorDict[5], nil)

    let phoneNumbers: ABMutableMultiValue =
    ABMultiValueCreateMutable(ABPropertyType(kABMultiStringPropertyType)).takeRetainedValue()
    ABMultiValueAddValueAndLabel(phoneNumbers, visitorDict["visitorPhone"], kABPersonPhoneMainLabel, nil)
    ABRecordSetValue(visitorRecord, kABPersonPhoneProperty, phoneNumbers, nil)

    ABAddressBookAddRecord(addressBookRef, visitorRecord, nil)
    saveAddressBookChanges()

    return visitorRecord
}

我喜欢触发

func addVisitorToContacts(sender: AnyObject) {
    //let visitor = ListVisitors[visitorButton.tag]
    var visitorDict:[Int:String] = [1:"\(visitorName)", 2:"\(visitorCompany)", 3:"\(visitorCity)",
        4:"\(visitorPhone)", 5:"\(visitorEmail)"]

    let visitorRecord: ABRecordRef = makeAndAddVisitorRecord2(visitorDict)
    let contactAddedAlert = UIAlertController(title: "\(visitorName) was successfully added.",
        message: nil, preferredStyle: .Alert)
    contactAddedAlert.addAction(UIAlertAction(title: "OK", style: .Cancel, handler: nil))
    presentViewController(contactAddedAlert, animated: true, completion: nil)
}

但是makeAndAddVisitorRecord2编译出错

 Cannot specialize non-generic type 'ABRecordRef' (aka 'AnyObject')

[编辑 1] 可行的解决方案但不是最佳的，因为我没有使用我的访问者结构

func makeAndAddVisitorRecord2(visitorDict: Dictionary <Int, String>) -> ABRecordRef  {

[编辑 2] 正如@rsmoz 指出的那样，我应该使用我的 Visitor 结构

class Visitor {

var visitorName : String
var visitorCompany : String
var visitorPlace : String
var visitorPhone : String
var visitorEmail : String

init(visitorName: String, visitorCompany: String, visitorPlace: String, visitorPhone: String, visitorEmail: String) {
    self.visitorName = visitorName
    self.visitorCompany = visitorCompany
    self.visitorPlace = visitorPlace
    self.visitorPhone = visitorPhone
    self.visitorEmail = visitorEmail
}

所以我有一个 ListVisitors 类，它生成一些 Visitors，看起来像

class ListVisitors{
    static var sharedInstance = [Visitor]()

static func load()
    {
        // @todo: stored and loaded data
        var visitor = Visitor(visitorName: "From Class Matt", visitorCompany: "Google", visitorPlace: "San Diego", visitorPhone: "94888484", visitorEmail: "matt@google.com")
        sharedInstance = [visitor]

        visitor = Visitor(visitorName: "From Class John", visitorCompany: "nike", visitorPlace: "New York", visitorPhone: "94888484", visitorEmail: "john@nike.com")
        //        ListVisitors.sharedInstance += [visitor]
        sharedInstance += [visitor]
...
}
}

在我的主 Controller 中，我有一个 TableView ，一个选定的行将访问者详细信息发送到 detailcontroller(我如何才能在详细 View Controller 中拥有选定的访问者结构？？我应该将 let selectedVisitor 传递给详细 View Controller 吗？)

override func prepareForSegue(segue: UIStoryboardSegue, sender: AnyObject?){
if (segue.identifier == "visitorDetails") {

            if let indexPath = tableView.indexPathForCell(sender as! UITableViewCell) {

                let selectedVisitor = lVisitors[indexPath.row] as Visitor

                let detailVC = segue.destinationViewController as! DetailViewController

                detailVC.visitorName = selectedVisitor.visitorName
                detailVC.visitorCompany = selectedVisitor.visitorCompany
                detailVC.visitorPlace = selectedVisitor.visitorPlace
                detailVC.visitorPhone = selectedVisitor.visitorPhone
                detailVC.visitorEmail = selectedVisitor.visitorEmail

            } // .end accessory select

        } // .end segue

Answer 1

我不确定你想用 ABRecordRef <AnyObject, AnyObject> 做什么，但是 <>语法用于指定通用类型。比如，一个包含字符串的数组是 Array<String> . ABRecordRef不是通用类型。

字典需要在参数中指定其包含的类型:Dictionary<String, Int>

此外，您将字典视为数组。最好使用字典，因为它应该被使用。而不是 [1:"\(visitorName)"] ，为什么不["visitorName":visitorName] ？这样你就可以像dict["visitorName"]一样访问它你也不需要做 "\(visitorName)"如果 visitorName 是一个字符串开头。直接使用变量即可。

不过，将访问者表示为结构而不是数组或字典会更好:

struct Visitor {
    let name: String
    let company: String
    let city: String
    let phone: String //Yes, this should be a String and not an Int
    let email: String
}

你可以这样设置:

let v = Visitor(name: "Joe", company: "A Corp", city: "New York", phone: "+44 392 39275 22", email: "joe@smith.org")

然后像这样访问它:

v.name

而且这样更清洁、更安全。现在，您的代码不会因意外访问字典中的错误键而出现任何错误。

哦，你应该使用 Contacts framework这些天，不是 ABAddressBook。