CLPlacemark 到 iOS 9 中的字符串

Question

我想将 CLPlacemark 格式化为字符串。

众所周知的方法是使用 ABCreateStringWithAddressDictionary 但它在 iOS 9 中已被弃用。警告告诉我改用 CNPostalAddressFormatter。

但是CNPostalAddressFormatter只能格式化CNPostalAddress。无法将 CLPlacemark 正确转换为 CNPostalAddress；只有这 3 个属性由 CLPlacemark 和 CNPostalAddress 共享：country、ISOcountryCode 和 postalCode.

那么我现在应该如何将 CLPlacemark 格式化为字符串？

Answer 1

获取地标的 addressDictionary 并使用其 "FormattedAddressLines" 键提取地址字符串。请注意，这是字符串行的数组。

（但是，您是对的，负责转换为联系人框架的 Apple 开发人员似乎完全忘记了地址簿和 CLPlacemark 之间的互换。这是联系人框架中的一个严重错误 - 许多错误之一.)

编辑因为我最初发布了那个答案，Apple 修复了这个错误。 CLPlacemark 现在有一个 postalAddress 属性，它是一个 CNPostalAddress，然后您可以使用 CNPostalAddressFormatter 来获得一个漂亮的多行地址字符串。务必import Contacts!

Answer 2

Swift 4.1（和 3 & 4，省 1 行）

我看了要问的问题'How might I implement this?':

extension String {
    init?(placemark: CLPlacemark?) {
        // Yadda, yadda, yadda
    }
}

两种方法

我和其他发帖人一样，首先着手移植 AddressDictionary 方法。但这意味着失去 CNPostalAddress class 和格式化程序的强大功能和灵活性。因此，方法 2.

extension String {
    // original method (edited)
    init?(depreciated placemark1: CLPlacemark?) {
    // UPDATE: **addressDictionary depreciated in iOS 11**
        guard
            let myAddressDictionary = placemark1?.addressDictionary,
            let myAddressLines = myAddressDictionary["FormattedAddressLines"] as? [String]
    else { return nil }

        self.init(myAddressLines.joined(separator: " "))
}

    // my preferred method - let CNPostalAddressFormatter do the heavy lifting
    init?(betterMethod placemark2: CLPlacemark?) {
        // where the magic is:
        guard let postalAddress = CNMutablePostalAddress(placemark: placemark2) else { return nil }
        self.init(CNPostalAddressFormatter().string(from: postalAddress))
    }
}

等等，CLPlacemark → CNPostalAddress 初始化器是什么？

extension CNMutablePostalAddress {
    convenience init(placemark: CLPlacemark) {
        self.init()
        street = [placemark.subThoroughfare, placemark.thoroughfare]
            .compactMap { [=12=] }           // remove nils, so that...
            .joined(separator: " ")      // ...only if both != nil, add a space.
    /*
    // Equivalent street assignment, w/o flatMap + joined:
        if let subThoroughfare = placemark.subThoroughfare,
            let thoroughfare = placemark.thoroughfare {
            street = "\(subThoroughfare) \(thoroughfare)"
        } else {
            street = (placemark.subThoroughfare ?? "") + (placemark.thoroughfare ?? "")
        } 
    */
        city = placemark.locality ?? ""
        state = placemark.administrativeArea ?? ""
        postalCode = placemark.postalCode ?? ""
        country = placemark.country ?? ""
        isoCountryCode = placemark.isoCountryCode ?? ""
        if #available(iOS 10.3, *) {
            subLocality = placemark.subLocality ?? ""
            subAdministrativeArea = placemark.subAdministrativeArea ?? ""
        }
    }
}

用法

func quickAndDirtyDemo() {
    let location = CLLocation(latitude: 38.8977, longitude: -77.0365)

    CLGeocoder().reverseGeocodeLocation(location) { (placemarks, _) in
        if let address = String(depreciated: placemarks?.first) {
            print("\nAddress Dictionary method:\n\(address)") }

        if let address = String(betterMethod: placemarks?.first) {
            print("\nEnumerated init method:\n\(address)") }
    }
}

/* Output:
Address Dictionary method:
The White House 1600 Pennsylvania Ave NW Washington, DC  20500 United States

Enumerated init method:
1600 Pennsylvania Ave NW
Washington DC 20500
United States
*/

读到这里的人将获得一件免费的 T 恤。（不是真的）

*此代码在 Swift 3 和 4 中有效，除了用于删除 nil 值的 flatMap 在 Swift 4.1 中已从 depreciated/renamed 到 compactMap （文档 here, or see SE-187 的基本原理）。

Answer 3

Swift 3.0

if let lines = myCLPlacemark.addressDictionary?["FormattedAddressLines"] as? [String] {
    let placeString = lines.joined(separator: ", ")
    // Do your thing
}

Answer 4

Swift 3.0 辅助方法

class func addressFromPlacemark(_ placemark:CLPlacemark)->String{
        var address = ""

        if let name = placemark.addressDictionary?["Name"] as? String {
            address = constructAddressString(address, newString: name)
        }

        if let city = placemark.addressDictionary?["City"] as? String {
            address = constructAddressString(address, newString: city)
        }

        if let state = placemark.addressDictionary?["State"] as? String {
            address = constructAddressString(address, newString: state)
        }

        if let country = placemark.country{
          address = constructAddressString(address, newString: country)
        }

        return address
      }

CLPlacemark 到 iOS 9 中的字符串

CLPlacemark to string in iOS 9

geocoding

objective-c

ios

swift

Swift 4.1（和 3 & 4，省 1 行）

我看了要问的问题'How might I implement this?':

两种方法

用法

Swift 3.0