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

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>

其他回答

你仍然可以在Swift中使用nsdictionary:

Swift 4

 var nsDictionary: NSDictionary?
 if let path = Bundle.main.path(forResource: "Config", ofType: "plist") {
    nsDictionary = NSDictionary(contentsOfFile: path)
 }

Swift 3+

if let path = Bundle.main.path(forResource: "Config", ofType: "plist"),
   let myDict = NSDictionary(contentsOfFile: path){
    // Use your myDict here
}

以及旧版本的Swift

var myDict: NSDictionary?
if let path = NSBundle.mainBundle().pathForResource("Config", ofType: "plist") {
    myDict = NSDictionary(contentsOfFile: path)
}
if let dict = myDict {
    // Use your dict here
}

NSClasses仍然可用,完全可以在Swift中使用。我想他们可能很快就会把重点转移到swift上,但是目前swift api并没有核心NSClasses的所有功能。

下面是一个基于@connor的回答的简短版本

guard let path = Bundle.main.path(forResource: "GoogleService-Info", ofType: "plist"),
    let myDict = NSDictionary(contentsOfFile: path) else {
    return nil
}

let value = dict.value(forKey: "CLIENT_ID") as! String?

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 {

  }
}

通过尼克的回答转换成一个方便的扩展:

extension Dictionary {
    static func contentsOf(path: URL) -> Dictionary<String, AnyObject> {
        let data = try! Data(contentsOf: path)
        let plist = try! PropertyListSerialization.propertyList(from: data, options: .mutableContainers, format: nil)

        return plist as! [String: AnyObject]
    }
}

用法:

let path = Bundle.main.path(forResource: "plistName", ofType: "plist")!
let url = URL(fileURLWithPath: path)
let dict = Dictionary<String, AnyObject>.contentsOf(path: url)

我敢打赌,它也可以为数组创建类似的扩展

这个答案使用Swift本机对象而不是NSDictionary。

斯威夫特3.0

//get the path of the plist file
guard let plistPath = Bundle.main.path(forResource: "level1", ofType: "plist") else { return }
//load the plist as data in memory
guard let plistData = FileManager.default.contents(atPath: plistPath) else { return }
//use the format of a property list (xml)
var format = PropertyListSerialization.PropertyListFormat.xml
//convert the plist data to a Swift Dictionary
guard let  plistDict = try! PropertyListSerialization.propertyList(from: plistData, options: .mutableContainersAndLeaves, format: &format) as? [String : AnyObject] else { return }
//access the values in the dictionary 
if let value = plistDict["aKey"] as? String {
  //do something with your value
  print(value)
}
//you can also use the coalesce operator to handle possible nil values
var myValue = plistDict["aKey"] ?? ""