我提供了另一个答案，因为这里没有一个答案是针对从测试包加载资源的。如果您正在使用一个输出JSON的远程服务，并且希望在不触及实际服务的情况下对解析结果进行单元测试，则可以获取一个或多个响应，并将它们放入项目中的Tests文件夹中的文件中。

func testCanReadTestJSONFile() {
    let path = NSBundle(forClass: ForecastIOAdapterTests.self).pathForResource("ForecastIOSample", ofType: "json")
    if let jsonData = NSData(contentsOfFile:path!) {
        let json = JSON(data: jsonData)
        if let currentTemperature = json["currently"]["temperature"].double {
            println("json: \(json)")
            XCTAssertGreaterThan(currentTemperature, 0)
        }
    }
}

这也使用了SwiftyJSON，但获得测试包和加载文件的核心逻辑是问题的答案。

根据Abhishek的回答，对于iOS 8，这将是:

let masterDataUrl: NSURL = NSBundle.mainBundle().URLForResource("masterdata", withExtension: "json")!
let jsonData: NSData = NSData(contentsOfURL: masterDataUrl)!
let jsonResult: NSDictionary = NSJSONSerialization.JSONObjectWithData(jsonData, options: nil, error: nil) as! NSDictionary
var persons : NSArray = jsonResult["person"] as! NSArray

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
    }

Swift 5.1, Xcode 11

你可以用这个:

struct Person : Codable {
    let name: String
    let lastName: String
    let age: Int
}

func loadJson(fileName: String) -> Person? {
   let decoder = JSONDecoder()
   guard
        let url = Bundle.main.url(forResource: fileName, withExtension: "json"),
        let data = try? Data(contentsOf: url),
        let person = try? decoder.decode(Person.self, from: data)
   else {
        return nil
   }

   return person
}

Swift 5的答案为我工作，除了我必须添加一个空文件，重命名为xxx。Json，并使用泛型。

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

code

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
    }
}