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
如果你有信息。Plist，然后使用
Bundle.main.infoDictionary
其他回答
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?
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)")
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 {
}
}
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
}
