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

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
}

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 {

  }
}

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。

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

因为这个答案还没有在这里,只是想指出,你也可以使用infoDictionary属性来获取信息plist作为字典,Bundle.main.infoDictionary。

比如bundle。main。如果您只对info plist中的特定项感兴趣,则可以使用kCFBundleNameKey作为String)。

// Swift 4

// Getting info plist as a dictionary
let dictionary = Bundle.main.infoDictionary

// Getting the app display name from the info plist
Bundle.main.infoDictionary?[kCFBundleNameKey as String]

// Getting the app display name from the info plist (another way)
Bundle.main.object(forInfoDictionaryKey: kCFBundleNameKey as String)

你可以使用它,我在github https://github.com/DaRkD0G/LoadExtension中创建了一个简单的字典扩展

extension Dictionary {
    /**
        Load a Plist file from the app bundle into a new dictionary

        :param: File name
        :return: Dictionary<String, AnyObject>?
    */
    static func loadPlistFromProject(filename: String) -> Dictionary<String, AnyObject>? {

        if let path = NSBundle.mainBundle().pathForResource("GameParam", ofType: "plist") {
            return NSDictionary(contentsOfFile: path) as? Dictionary<String, AnyObject>
        }
        println("Could not find file: \(filename)")
        return nil
    }
}

你可以用它来计算负载

/**
  Example function for load Files Plist

  :param: Name File Plist
*/
func loadPlist(filename: String) -> ExampleClass? {
    if let dictionary = Dictionary<String, AnyObject>.loadPlistFromProject(filename) {
        let stringValue = (dictionary["name"] as NSString)
        let intergerValue = (dictionary["score"] as NSString).integerValue
        let doubleValue = (dictionary["transition"] as NSString).doubleValue

        return ExampleClass(stringValue: stringValue, intergerValue: intergerValue, doubleValue: doubleValue)
    }
    return nil
}