我在玩苹果的新Swift编程语言,遇到了一些问题…

目前我试图读取一个plist文件,在Objective-C中,我会做以下工作来获取内容作为NSDictionary:

NSString *filePath = [[NSBundle mainBundle] pathForResource:@"Config" ofType:@"plist"];
NSDictionary *dict = [[NSDictionary alloc] initWithContentsOfFile:filePath];

我如何得到一个plist作为一个字典在Swift?

我假设我可以得到路径到plist:

let path = NSBundle.mainBundle().pathForResource("Config", ofType: "plist")

当这工作(如果它是正确的?):我如何获得内容作为一个字典?

还有一个更普遍的问题:

是否可以使用默认的NS*类?我想是的……还是我遗漏了什么?据我所知,默认框架NS*类仍然有效,可以使用吗?


当前回答

访问plist文件的简单结构(Swift 2.0)

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

  static let someValue = dict["someKey"] as! String
}

用法:

print("someValue = \(Configuration.someValue)")

其他回答

在我的情况下,我创建了一个NSDictionary称为appSettings并添加所有需要的键。对于这种情况,解决方案是:

if let dict = NSBundle.mainBundle().objectForInfoDictionaryKey("appSettings") {
  if let configAppToken = dict["myKeyInsideAppSettings"] as? String {

  }
}

在SWIFT语言中,你可以这样读plist:

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

读单个字典值:

let test: AnyObject = dict.objectForKey("index1")

如果你想在plist中获得完整的多维字典:

let value: AnyObject = dict.objectForKey("index2").objectForKey("date")

下面是plist:

<plist version="1.0">
<dict>
<key>index2</key>
<dict>
    <key>date</key>
    <string>20140610</string>
    <key>amount</key>
    <string>110</string>
</dict>
<key>index1</key>
<dict>
    <key>amount</key>
    <string>125</string>
    <key>date</key>
    <string>20140212</string>
</dict>
</dict>
</plist>

访问plist文件的简单结构(Swift 2.0)

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

  static let someValue = dict["someKey"] as! String
}

用法:

print("someValue = \(Configuration.someValue)")

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").

斯威夫特4.0

现在可以使用decodedable协议将.plist解码为自定义结构。我将介绍一个基本的例子,对于更复杂的。plist结构,我建议阅读Decodable/Encodable(一个很好的资源是:https://benscheirman.com/2017/06/swift-json/)。

首先将结构设置为.plist文件的格式。对于这个例子,我将考虑一个根级字典和3个条目:1个字符串键“name”,1个Int键“age”,1个布尔键“single”。下面是结构体:

struct Config: Decodable {
    private enum CodingKeys: String, CodingKey {
        case name, age, single
    }

    let name: String
    let age: Int
    let single: Bool
}

很简单。现在是最酷的部分。使用PropertyListDecoder类,我们可以很容易地将.plist文件解析为这个结构体的实例化:

func parseConfig() -> Config {
    let url = Bundle.main.url(forResource: "Config", withExtension: "plist")!
    let data = try! Data(contentsOf: url)
    let decoder = PropertyListDecoder()
    return try! decoder.decode(Config.self, from: data)
}

不用担心太多代码,而且都在Swift中。更好的是,我们现在有一个Config结构的实例化,我们可以很容易地使用:

let config = parseConfig()
print(config.name) 
print(config.age)
print(config.single) 

打印。plist中“name”、“age”和“single”键的值。