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
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。
其他回答
能在一行内完成吗
var dict = NSDictionary(contentsOfFile: NSBundle.mainBundle().pathForResource("Config", ofType: "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
}
如果你有信息。Plist，然后使用
Bundle.main.infoDictionary
步骤1:简单和最快的方法来解析plist在swift 3+
extension Bundle {
func parsePlist(ofName name: String) -> [String: AnyObject]? {
// check if plist data available
guard let plistURL = Bundle.main.url(forResource: name, withExtension: "plist"),
let data = try? Data(contentsOf: plistURL)
else {
return nil
}
// parse plist into [String: Anyobject]
guard let plistDictionary = try? PropertyListSerialization.propertyList(from: data, options: [], format: nil) as? [String: AnyObject] else {
return nil
}
return plistDictionary
}
}
第二步:使用方法:
Bundle().parsePlist(ofName: "Your-Plist-Name")
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)。
