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?
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
}
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
}
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。
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)。