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。

简化Peter Kreinz提供的例子。适用于Swift 4.2。

扩展函数:

extension Decodable {
  static func parse(jsonFile: String) -> Self? {
    guard let url = Bundle.main.url(forResource: jsonFile, withExtension: "json"),
          let data = try? Data(contentsOf: url),
          let output = try? JSONDecoder().decode(self, from: data)
        else {
      return nil
    }

    return output
  }
}

示例模型:

struct Service: Decodable {
  let name: String
}

示例用法:

/// service.json
/// { "name": "Home & Garden" }

guard let output = Service.parse(jsonFile: "service") else {
// do something if parsing failed
 return
}

// use output if all good

这个例子也适用于数组:

/// services.json
/// [ { "name": "Home & Garden" } ]

guard let output = [Service].parse(jsonFile: "services") else {
// do something if parsing failed
 return
}

// use output if all good

注意，我们没有提供任何不必要的泛型，因此不需要强制转换parse的结果。

//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)")
    }
}

遵循以下代码:

if let path = NSBundle.mainBundle().pathForResource("test", ofType: "json")
{
    if let jsonData = NSData(contentsOfFile: path, options: .DataReadingMappedIfSafe, error: nil)
    {
        if let jsonResult: NSDictionary = NSJSONSerialization.JSONObjectWithData(jsonData, options: NSJSONReadingOptions.MutableContainers, error: nil) as? NSDictionary
        {
            if let persons : NSArray = jsonResult["person"] as? NSArray
            {
                // Do stuff
            }
        }
     }
}

数组“persons”将包含关键人物的所有数据。遍历获取它。

斯威夫特4.0:

if let path = Bundle.main.path(forResource: "test", ofType: "json") {
    do {
          let data = try Data(contentsOf: URL(fileURLWithPath: path), options: .mappedIfSafe)
          let jsonResult = try JSONSerialization.jsonObject(with: data, options: .mutableLeaves)
          if let jsonResult = jsonResult as? Dictionary<String, AnyObject>, let person = jsonResult["person"] as? [Any] {
                    // do stuff
          }
      } catch {
           // handle error
      }
}

下面的代码适用于我。我用的是Swift 5

let path = Bundle.main.path(forResource: "yourJSONfileName", ofType: "json")
var jsonData = try! String(contentsOfFile: path!).data(using: .utf8)!

然后，如果你的Person结构(或类)是可解码的(以及它的所有属性)，你可以简单地做:

let person = try! JSONDecoder().decode(Person.self, from: jsonData)

我避免了所有的错误处理代码，使代码更容易读懂。

最新的swift 3.0绝对有效

func loadJson(filename fileName: String) -> [String: AnyObject]?
{
    if let url = Bundle.main.url(forResource: fileName, withExtension: "json") 
{
      if let data = NSData(contentsOf: url) {
          do {
                    let object = try JSONSerialization.jsonObject(with: data as Data, options: .allowFragments)
                    if let dictionary = object as? [String: AnyObject] {
                        return dictionary
                    }
                } catch {
                    print("Error!! Unable to parse  \(fileName).json")
                }
            }
            print("Error!! Unable to load  \(fileName).json")
        }
        return nil
    }