在swift 3.0从Plist读取。

func readPropertyList() {
        var propertyListFormat =  PropertyListSerialization.PropertyListFormat.xml //Format of the Property List.
        var plistData: [String: AnyObject] = [:] //Our data
        let plistPath: String? = Bundle.main.path(forResource: "data", ofType: "plist")! //the path of the data
        let plistXML = FileManager.default.contents(atPath: plistPath!)!
        do {//convert the data to a dictionary and handle errors.
            plistData = try PropertyListSerialization.propertyList(from: plistXML, options: .mutableContainersAndLeaves, format: &propertyListFormat) as! [String:AnyObject]

        } catch {
            print("Error reading plist: \(error), format: \(propertyListFormat)")
        }
    }

阅读更多 如何在swift中使用属性列表(. plist)。

如果我想将.plist转换为Swift字典，这是我所做的:

if let path = NSBundle.mainBundle().pathForResource("Config", ofType: "plist") {
  if let dict = NSDictionary(contentsOfFile: path) as? Dictionary<String, AnyObject> {
    // use swift dictionary as normal
  }
}

为Swift 2.0编辑:

if let path = NSBundle.mainBundle().pathForResource("Config", ofType: "plist"), dict = NSDictionary(contentsOfFile: path) as? [String: AnyObject] {
    // use swift dictionary as normal
}

为Swift 3.0编辑:

if let path = Bundle.main.path(forResource: "Config", ofType: "plist"), let dict = NSDictionary(contentsOfFile: path) as? [String: AnyObject] {
        // use swift dictionary as normal
}

最好使用本机字典和数组，因为它们已经针对swift进行了优化。也就是说你可以使用NS。我认为这种情况证明了这一点。下面是你如何实现它:

var path = NSBundle.mainBundle().pathForResource("Config", ofType: "plist")
var dict = NSDictionary(contentsOfFile: path)

到目前为止(在我看来)，这是访问plist最简单和最有效的方法，但在未来，我希望苹果将添加更多的功能(比如使用plist)到本地字典中。

我已经创建了一个简单的字典初始化器替换NSDictionary(contentsOfFile: path)。只要去掉NS。

extension Dictionary where Key == String, Value == Any {

    public init?(contentsOfFile path: String) {
        let url = URL(fileURLWithPath: path)

        self.init(contentsOfURL: url)
    }

    public init?(contentsOfURL url: URL) {
        guard let data = try? Data(contentsOf: url),
            let dictionary = (try? PropertyListSerialization.propertyList(from: data, options: [], format: nil) as? [String: Any]) ?? nil
            else { return nil }

        self = dictionary
    }

}

你可以这样使用它:

let filePath = Bundle.main.path(forResource: "Preferences", ofType: "plist")!
let preferences = Dictionary(contentsOfFile: filePath)!
UserDefaults.standard.register(defaults: preferences)

斯威夫特5

如果你想获取某个键的特定值，那么我们可以使用下面的扩展，它使用Bundle上的infoDictionary属性。

可以使用Bundle.main.infoDictionary获取所有信息。在表单字典中plist值，因此我们可以直接使用Bundle上的object(forInfoDictionaryKey: key)方法查询

extension Bundle {
    static func infoPlistValue(forKey key: String) -> Any? {
        guard let value = Bundle.main.object(forInfoDictionaryKey: key) else {
           return nil
        }
        return value
    }
}

使用

guard let apiURL = Bundle.infoPlistValue(forKey: "API_URL_KEY") as? String else { return }

Swift 2.0:访问信息。Plist

我有一个名为CoachMarksDictionary的字典，在信息中具有布尔值。Plist。我想访问bool值并使其为真。

let path = NSBundle.mainBundle().pathForResource("Info", ofType: "plist")!
  let dict = NSDictionary(contentsOfFile: path) as! [String: AnyObject]

  if let CoachMarksDict = dict["CoachMarksDictionary"] {
       print("Info.plist : \(CoachMarksDict)")

   var dashC = CoachMarksDict["DashBoardCompleted"] as! Bool
    print("DashBoardCompleted state :\(dashC) ")
  }

写给Plist:

自定义Plist:- Make from File-New-File-Resource-PropertyList。增加了三个字符串:DashBoard_New, DashBoard_Draft, DashBoard_Completed)

func writeToCoachMarksPlist(status:String?,keyName:String?)
 {
  let path1 = NSBundle.mainBundle().pathForResource("CoachMarks", ofType: "plist")
  let coachMarksDICT = NSMutableDictionary(contentsOfFile: path1!)! as NSMutableDictionary
  var coachMarksMine = coachMarksDICT.objectForKey(keyName!)

  coachMarksMine  = status
  coachMarksDICT.setValue(status, forKey: keyName!)
  coachMarksDICT.writeToFile(path1!, atomically: true)
 }

该方法可以调用为

self.writeToCoachMarksPlist(" true - means user has checked the marks",keyName: "the key in the CoachMarks dictionary").