Question

尝试从AddressDictionary获取格式化地址，这是我从CLGeocoder获得的。使用以下代码，但没有结果：

subtitle = [NSString stringWithString:[[addressDict objectForKey:@"FormattedAddressLines"]objectAtIndex:0]];

也尝试过：

subtitle = [[[ABAddressBook sharedAddressBook] formattedAddressFromDictionary:placemark.addressDictionary] string];

但此代码似乎仅适用于Mac OS X.

编译器询问ABAdressBook，但我导入了两个头文件。

#import <AddressBook/ABAddressBook.h>
#import <AddressBook/AddressBook.h>

Answer 1

addressDictionary属性的文档说：

您可以格式化此词典的内容以获取完整地址字符串而不是自己构建地址。格式化字典，使用ABCreateStringWithAddressDictionary函数作为在地址簿UI函数参考中描述。

所以添加并导入AddressBookUI框架并尝试：

subtitle = 
    ABCreateStringWithAddressDictionary(placemark.addressDictionary, NO);

Answer 2

在iOS 6.1下进行一些挖掘后，我发现CLPlacemark地址字典包含一个预先格式化的地址：

CLLocation *location = [[CLLocation alloc]initWithLatitude:37.3175 longitude:-122.041944];
[[[CLGeocoder alloc]init] reverseGeocodeLocation:location completionHandler:^(NSArray *placemarks, NSError *error) {
    CLPlacemark *placemark = placemarks[0];
    NSArray *lines = placemark.addressDictionary[ @"FormattedAddressLines"];
    NSString *addressString = [lines componentsJoinedByString:@"\n"];
    NSLog(@"Address: %@", addressString);
}];

我还没有找到关于此的文档，但它适用于我测试过的所有地址。

Answer 3

正如Martyn Davis强调的那样，{9}中不推荐使用ABCreateStringWithAddressDictionary。

您可以使用以下功能将addressDictionary转换为较新的CNMutablePostalAddress，然后使用CNPostalAddressFormatter生成本地化字符串，只要您导入Contacts即可框架。

Swift 3.x

// Convert to the newer CNPostalAddress
func postalAddressFromAddressDictionary(_ addressdictionary: Dictionary<NSObject,AnyObject>) -> CNMutablePostalAddress {
   let address = CNMutablePostalAddress()

   address.street = addressdictionary["Street" as NSObject] as? String ?? ""
   address.state = addressdictionary["State" as NSObject] as? String ?? ""
   address.city = addressdictionary["City" as NSObject] as? String ?? ""
   address.country = addressdictionary["Country" as NSObject] as? String ?? ""
   address.postalCode = addressdictionary["ZIP" as NSObject] as? String ?? ""

   return address
}

// Create a localized address string from an Address Dictionary
func localizedStringForAddressDictionary(addressDictionary: Dictionary<NSObject,AnyObject>) -> String {
    return CNPostalAddressFormatter.string(from: postalAddressFromAddressDictionary(addressDictionary), style: .mailingAddress)
}

Swift 2.x

import Contacts

// Convert to the newer CNPostalAddress
func postalAddressFromAddressDictionary(addressdictionary: Dictionary<NSObject,AnyObject>) -> CNMutablePostalAddress {

    let address = CNMutablePostalAddress()

    address.street = addressdictionary["Street"] as? String ?? ""
    address.state = addressdictionary["State"] as? String ?? ""
    address.city = addressdictionary["City"] as? String ?? ""
    address.country = addressdictionary["Country"] as? String ?? ""
    address.postalCode = addressdictionary["ZIP"] as? String ?? ""

    return address
}

// Create a localized address string from an Address Dictionary
func localizedStringForAddressDictionary(addressDictionary: Dictionary<NSObject,AnyObject>) -> String {

    return CNPostalAddressFormatter.stringFromPostalAddress(postalAddressFromAddressDictionary(addressDictionary), style: .MailingAddress)
}

Answer 4

func locationManager(manager: CLLocationManager, didUpdateLocations locations: [CLLocation]) {
    // get the address
    if let location = locations.last {
        CLGeocoder().reverseGeocodeLocation(location, completionHandler: { (result: [CLPlacemark]?, err: NSError?) -> Void in
            if let placemark = result?.last
                , addrList = placemark.addressDictionary?["FormattedAddressLines"] as? [String]
            {
                let address =  addrList.joinWithSeparator(", ")
                print(address)
            }
        })
    }
}

以上是快速版本。

Answer 5

我正在使用Swift 3 / XCode 8

ZYiOS's answer很好而且简短，但没有为我编译。

问题是如何从现有的地址字典到字符串地址。这就是我所做的：

import CoreLocation

func getAddressString(placemark: CLPlacemark) -> String? {
    var originAddress : String?

    if let addrList = placemark.addressDictionary?["FormattedAddressLines"] as? [String]
    {
        originAddress =  addrList.joined(separator: ", ")
    }

    return originAddress
}

Answer 6

Swift 3 / Xcode 8 帮助Mehtod从CLPlaceMark获取地址

class func formattedAddress(fromPlacemark 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
  }

Answer 7

现在这很简单

UIPasteboard

Answer 8

只需为CLLocation创建扩展程序：

typealias AddressDictionaryHandler = ([String: Any]?) -> Void

extension CLLocation {

    func addressDictionary(completion: @escaping AddressDictionaryHandler) {

        CLGeocoder().reverseGeocodeLocation(self) { placemarks, _ in
            completion(placemarks?.first?.addressDictionary as? [String: AnyObject])
        }
    }
}

示例：

let location = CLLocation()

location.addressDictionary { dictionary in

    let city = dictionary?["City"] as? String
    let street = dictionary?["Street"] as? String
}

Answer 9

Swift 5版本

onLeave

Answer 10

iOS 11+

import CoreLocation
import Contacts

public extension CLPlacemark {
    func formattedAddress() -> String? {
        guard let postalAddress = postalAddress else { return nil }
        let formatter = CNPostalAddressFormatter()
        formatter.style = .mailingAddress
        let formatterString = formatter.string(from: postalAddress)
        return formatterString.replacingOccurrences(of: "\n", with: " ")
    }
}

如何从AddressDictionary获取格式化地址NSString？

10 个答案:

Swift 3.x

Swift 2.x