获取ABPeoplePickerView的分离值

Question

如何获得我在ABPeoplePickerView中选择的人的地址值(城市、街道、拉链)？

   let person = PeoplePicker.selectedRecords as! [ABPerson]

    print(person[0].value(forProperty: "First"))
    print(person[0].value(forProperty: "Last"))
    print(person[0].value(forProperty: "Address"))

这是我的打印结果：

Max
Mustermann
{
        *  home  {
    City = xxx;
    Street = "xxx";
    ZIP = xxx;
}  D
}

现在编码：

let addressBook = ABAddressBook.shared()
        let people = addressBook?.people()
        for person in people! as! [ABPerson] {
            for property in ABPerson.properties() {
                if let multiValue = person.value(forProperty: (property as! NSString) as String!) as? ABMultiValue {
                    for identifier in multiValue {
                        let value: AnyObject = multiValue.valueForIdentifier(identifier as String)
                        print("\(identifier) : \(value)")
                    }
                }
            }
        }

错误：

Type 'ABMultiValue' does not conform to protocol 'Sequence'
on line: for identifier in multiValue {

Answer 1

解决办法是：

let address = person[0].value(forProperty: kABAddressProperty) as! ABMultiValue
let zip = address.value(at: 0) as! NSMutableDictionary
print(y.value(forKey: kABAddressZIPKey))

Answer 2

参考此示例，并对单个联系人进行修改。

let addressBook = ABAddressBook.sharedAddressBook()
let people = addressBook.people()
for person in people {
    for property in ABPerson.properties() {
        if let multiValue = person.valueForProperty(property as NSString) as? ABMultiValue {
            for identifier in multiValue {
                let value: AnyObject = multiValue.valueForIdentifier(identifier as String)
                println("\(identifier) : \(value)")
            }
        }
    }
}

Answer 3

Get值是一个带有邮政地址值的标记值数组。邮政地址有“街道”财产、“城市”财产、“国家”财产、“postalCode”财产、“国家”财产。

let array = person[0].value(forProperty: "Address") as! NSarray

使用标签值获取该值，您可以获得像城市、州和邮政编码这样分开的名称。