我真的在努力把一个JSON文件读入Swift,这样我就可以玩它了。我花了2天的时间重新搜索和尝试不同的方法,但没有运气,所以我已经注册了StackOverFlow,看看是否有人能给我指点正确的方向.....

我的JSON文件叫做test。Json,并包含以下内容:

{
  "person":[
     {
       "name": "Bob",
       "age": "16",
       "employed": "No"
     },
     {
       "name": "Vinny",
       "age": "56",
       "employed": "Yes"
     }
  ]
}    

该文件直接存储在文档中,我使用以下代码访问它:

let file = "test.json"
let dirs : String[] = NSSearchPathForDirectoriesInDomains(
                                                          NSSearchpathDirectory.DocumentDirectory,
                                                          NSSearchPathDomainMask.AllDomainMask,
                                                          true) as String[]

if (dirs != nil) {
    let directories: String[] = dirs
    let dir = directories[0]
    let path = dir.stringByAppendingPathComponent(file)
}

var jsonData = NSData(contentsOfFile:path, options: nil, error: nil)
println("jsonData \(jsonData)" // This prints what looks to be JSON encoded data.

var jsonDict = NSJSONSerialization.JSONObjectWithData(jsonData, options: nil, error: nil) as? NSDictionary

println("jsonDict \(jsonDict)") - This prints nil..... 

如果有人能给我一个正确的方向,我可以反序列化JSON文件,并把它放在一个可访问的Swift对象,我会永远感激!

亲切的问候,

Krivvenz。


当前回答

在清理和抛光我的代码之后,我来到了这两个函数,你可以添加到你的项目中,并使用它们非常整洁和快速地从json文件读取数据,并将数据转换为你想要的任何类型!

public func readDataRepresentationFromFile(resource: String, type: String) -> Data? {
    let filePath = Bundle.main.path(forResource: resource, ofType: type)
    
    if let path = filePath {
        let result = FileManager.default.contents(atPath: path)
        return result
    }
    return nil
}

然后在这个函数的帮助下,你可以将你的数据转换为任何你想要的类型:

public func getObject<T: Codable>(of type: T.Type, from file: String) -> T?  {
    guard let data = readDataRepresentationFromFile(resource: file, type: "json") else {
        return nil
    }
    if let object = try? JSONDecoder().decode(type, from: data) {
        return object
    }
    return nil
}

此代码的应用示例: 在你的代码中调用这个函数,给它你的json文件的名字,这就是你所需要的!

func getInputDataFromSomeJson(jsonFileName: String) -> YourReqiuredOutputType? {
    return getObject(of: YourReqiuredOutputType.self, from: jsonFileName)
}

其他回答

我可能还会推荐Ray Wenderlich的Swift JSON教程(它还涵盖了很棒的SwiftyJSON替代品,Gloss)。一段摘录(它本身并不能完全回答海报上的问题,但这个答案的附加价值是链接,所以请不要给它加-1):

在Objective-C中,解析和反序列化JSON相当简单:

NSArray *json = [NSJSONSerialization JSONObjectWithData:JSONData
options:kNilOptions error:nil];
NSString *age = json[0][@"person"][@"age"];
NSLog(@"Dani's age is %@", age);

在Swift中,由于Swift的可选选项和类型安全,解析和反序列化JSON有点繁琐,但作为Swift 2.0的一部分,guard语句被引入,以帮助摆脱嵌套的if语句:

var json: Array!
do {
  json = try NSJSONSerialization.JSONObjectWithData(JSONData, options: NSJSONReadingOptions()) as? Array
} catch {
  print(error)
}

guard let item = json[0] as? [String: AnyObject],
  let person = item["person"] as? [String: AnyObject],
  let age = person["age"] as? Int else {
    return;
}
print("Dani's age is \(age)")

当然,在XCode 8中。x,你只需双击空格键,然后说“嘿,Siri,请在Swift 3.0中用空格/制表符缩进为我反序列化这个JSON。”

Swift 4:试试我的解决方案:

test.json

{
    "person":[
        {
            "name": "Bob",
            "age": "16",
            "employed": "No"
        },
        {
            "name": "Vinny",
            "age": "56",
            "employed": "Yes"
        }
    ]
}

RequestCodable.swift

import Foundation

struct RequestCodable:Codable {
    let person:[PersonCodable]
}

PersonCodable.swift

import Foundation

struct PersonCodable:Codable {
    let name:String
    let age:String
    let employed:String
}

可解码+ FromJSON.swift

import Foundation

extension Decodable {

    static func fromJSON<T:Decodable>(_ fileName: String, fileExtension: String="json", bundle: Bundle = .main) throws -> T {
        guard let url = bundle.url(forResource: fileName, withExtension: fileExtension) else {
            throw NSError(domain: NSURLErrorDomain, code: NSURLErrorResourceUnavailable)
        }

        let data = try Data(contentsOf: url)

        return try JSONDecoder().decode(T.self, from: data)
    }
}

例子:

let result = RequestCodable.fromJSON("test") as RequestCodable?

result?.person.compactMap({ print($0) }) 

/*
PersonCodable(name: "Bob", age: "16", employed: "No")
PersonCodable(name: "Vinny", age: "56", employed: "Yes")
*/
//change type based on your struct and right JSON file

let quoteData: [DataType] =
    load("file.json")

func load<T: Decodable>(_ filename: String, as type: T.Type = T.self) -> T {
    let data: Data

    guard let file = Bundle.main.url(forResource: filename, withExtension: nil)
        else {
            fatalError("Couldn't find \(filename) in main bundle.")
    }

    do {
        data = try Data(contentsOf: file)
    } catch {
        fatalError("Couldn't load \(filename) from main bundle:\n\(error)")
    }

    do {
        let decoder = JSONDecoder()
        return try decoder.decode(T.self, from: data)
    } catch {
        fatalError("Couldn't parse \(filename) as \(T.self):\n\(error)")
    }
}


一般的方法可以是这样的:

创建响应类名称字符串的json文件

struct Response: Codable,FileDecodable {
    typealias T = Self
    let names:[Data]
}
protocol FileDecodable{
   associatedtype T:Codable
   static func loadJson() ->T?
}

extension FileDecodable{
    static func loadJson() -> T? {
        let fileName = String(describing: T.self)
        if let url = Bundle.main.url(forResource: fileName, withExtension: "json")     {
            do {
                let data = try Data(contentsOf: url)
                let decoder = JSONDecoder()
                let jsonData = try decoder.decode(T.self, from: data)
                return jsonData
            } catch {
                print("error:\(error)")
            }
        }
        return nil
    }
}

斯威夫特 5+

用Struct解码jsonData

if let jsonData = readFile(forName: <your file name>) {

do {
                let decodedData = try JSONDecoder().decode(<your struct name>.self, from: jsonData)
                return decodedData.<what you expect>
            } catch { print("JSON decode error") }
}

这将读取文件并返回jsonData

如果你实际上在另一个bundle中(例如test),使用: let bundlePath = Bundle(for: type(of: self))。路径(forResource: name, ofType: "json")

private func readFile(forName name: String) -> Data? {
        do {

            if let bundlePath = Bundle.main.path(forResource: name, ofType: "json"),
                let jsonData = try String(contentsOfFile: bundlePath).data(using: .utf8) {
                return jsonData
            }
        } catch {
            print(error)
        }
        return nil
    }