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")
*/

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

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
    }

对于那些希望在decodedable抛出错误后在调试器/ lldb中打印JSON数据的人，请尝试以下方法。

阿宝试试!JSONSerialization。jsonObject(:数据!，选项:.mutableLeaves) as?字典<字符串,任何>

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。

这里还有一个答案??

好的。坚持住!之前所有的答案都是关于使用JSONSerialization，或返回nil，或忽略错误。

有什么不同

“我的解决方案”(不是真的我的解决方案，这是上述解决方案的混合)包含:

返回值的现代方法:Result<Value,Error>(返回值或错误) 避免使用空值 包含一个稍微详细的错误 使用扩展有漂亮/直观的界面: 提供了选择包的可能性

细节

Xcode 14 斯威夫特5.6.1

解决方案1。JSON文件->可解码

enum JSONParseError: Error {
    case fileNotFound
    case dataInitialisation(error: Error)
    case decoding(error: Error)
}

extension Decodable {
    static func from(localJSON filename: String,
                     bundle: Bundle = .main) -> Result<Self, JSONParseError> {
        guard let url = bundle.url(forResource: filename, withExtension: "json") else {
            return .failure(.fileNotFound)
        }
        let data: Data
        do {
            data = try Data(contentsOf: url)
        } catch let error {
            return .failure(.dataInitialisation(error: error))
        }

        do {
            return .success(try JSONDecoder().decode(self, from: data))
        } catch let error {
            return .failure(.decoding(error: error))
        }
    }
}

方案一用途

 struct Model: Decodable {
    let uuid: String
    let name: String
}

switch Model.from(localJSON: "myjsonfile") {
case .success(let value):
    print(value)
case .failure(let error):
    print(error)
}

解决方案2。JSON文件->字典

extension Dictionary where Key == String, Value == Any {

    enum JSONParseError: Error {
        case fileNotFound(filename: String)
        case dataInitialisation(Error)
        case jsonSerialization(Error)
        case mappingFail(value: Any, toType: Any)
    }

    static func from(JSONfile url: URL) -> Result<Self, JSONParseError> {
        let data: Data
        do {
            data = try Data(contentsOf: url)
        } catch let error {
            return .failure(.dataInitialisation(error))
        }

        let jsonObject: Any
        do {
            jsonObject = try JSONSerialization.jsonObject(with: data, options: .mutableLeaves)
        } catch let error {
            return .failure(.jsonSerialization(error))
        }

        guard let jsonResult = jsonObject as? Self else {
            return .failure(.mappingFail(value: jsonObject, toType: Self.Type.self))
        }

        return .success(jsonResult)
    }

    static func from(localJSONfile name: String) -> Result<Self, JSONParseError> {
        let fileType = "json"
        let fullFileName = name + (name.contains(fileType) ? "" : ".\(fileType)")
        guard let path = Bundle.main.path(forResource: fullFileName, ofType: "") else {
            return .failure(.fileNotFound(filename: fullFileName))
        }
        return from(JSONfile: URL(fileURLWithPath: path))
    }
}

方案二使用

switch [String: Any].from(localJSONfile: "file.json") {
// OR switch [String: Any].from(localJSONfile: "file.json") {
// OR switch [String: Any].from(JSONfile: url) {
case let .success(dictionary):
    print(dictionary)
case let .failure(error):
    print("ERROR: \(error)")
}