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
}
我想通过
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makeAndAddVisitorRecord2编译错误
Cannot specialize non-generic type 'ABRecordRef' (aka 'AnyObject')
[编辑1]可行的解决方案,但不是最优的,因为我没有使用我的访客结构
func makeAndAddVisitorRecord2(visitorDict: Dictionary <Int, String>) -> ABRecordRef {
[编辑2]正如@rsmoz所指出我应该使用我的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,它会生成一些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]
...
}
}
在我的主控制器中,我有一个表视图,并且选定的行将访问者详细信息发送到detailcontroller(我如何在详细视图控制器中拥有所选的访问者结构?我应该将selectVisitor传递给详细视图控制器吗?)
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
答案 0 :(得分:9)
我不确定您要对ABRecordRef <AnyObject, AnyObject>
做什么,但<>
语法用于指定泛型类型。就像,一个包含字符串的数组是Array<String>
。 ABRecordRef
不是通用类型。
词典需要具有在参数中指定的类型:Dictionary<String, Int>
另外,你正在处理类似数组的字典。最好使用字典,因为它意味着使用。而不是[1:"\(visitorName)"]
,为什么不["visitorName":visitorName]
?这样你可以像dict["visitorName"]
那样访问它。如果visitorName是一个开头的字符串,你也不需要"\(visitorName)"
。只需直接使用变量即可。
但是,将Visitor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:
v.name
而且那更加清洁和安全。现在,您的代码不会因意外地访问字典上的错误密钥而出现任何错误。
哦,你现在应该使用Contacts framework,而不是ABAddressBook。
答案 1 :(得分:4)
Swift中的字典类型称为Dictionary
。但是,它是通用的,这意味着您需要在类型名称后添加<>
以指定您希望它的字典类型。在这种情况下,它是Dictionary<Int, String>
。这是因为您传递给方法(visitorDict
)的变量属于Dictionary<Int, String>
类型。
像这样编写函数头:
func makeAndAddVisitorRecord2(visitorDict: Dictionary<Int, String>) -> ABRecordRef <AnyObject, AnyObject> {
如果您想更进一步,可以使用字典的简写类型名称:
func makeAndAddVisitorRecord2(visitorDict: [Int: String]) -> ABRecordRef <AnyObject, AnyObject> {
此外,如果您的字典键是顺序的,如1,2,3,4,5等,您可以使用数组:
func makeAndAddVisitorRecord2(visitorDict: Array<String>) -> ABRecordRef <AnyObject, AnyObject> {
数组的简写是:
func makeAndAddVisitorRecord2(visitorDict: [String]) -> ABRecordRef <AnyObject, AnyObject> {
答案 2 :(得分:1)
你必须在swift中指定它是什么类型的字典,如下所示:
func makeAndAddVisitorRecord2(visitorDict: [Int: String]) -> 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
}
然后你可以使用makeAndAddVisitorRecord2(visitorDict)
,一切都应该有效。
您的集合类型apple doc有更深入的示例。
答案 3 :(得分:0)
ABRecordRef不是通用的,因此你不能像你那样使用。您很可能想要将通用声明添加到Dictionary参数。