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

Swift 3.0的更新名称

根据阿比谢克和德鲁瓦的回答

func loadJson(forFilename fileName: String) -> NSDictionary? {

    if let url = Bundle.main.url(forResource: fileName, withExtension: "json") {
        if let data = NSData(contentsOf: url) {
            do {
                let dictionary = try JSONSerialization.jsonObject(with: data as Data, options: .allowFragments) as? NSDictionary

                return dictionary
            } catch {
                print("Error!! Unable to parse  \(fileName).json")
            }
        }
        print("Error!! Unable to load  \(fileName).json")
    }

    return nil
}

其他回答

使用这个泛型函数

func readJSONFromFile<T: Decodable>(fileName: String, type: T.Type) -> T? {
    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
}

下面这行代码:

let model = readJSONFromFile(fileName: "Model", type: Model.self)

对于这种类型:

struct Model: Codable {
    let tall: Int
}

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)")
}

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。

对于那些正在寻找可重用函数的人,我做了一个负责JSON加载的类。

import Foundation

class JSONLoader {
    static func load<T: Decodable>(resource: String, type: T.Type) -> T {
        guard let file = Bundle.main.url(forResource: resource, withExtension: nil) else {
            fatalError("Couldn't find \(resource) in main bundle.")
        }
        let data: Data
        do {
            data = try Data(contentsOf: file)
        } catch {
            fatalError("Couldn't load \(resource) from main bundle:\n\(error)")
        }
        do {
            return try JSONDecoder().decode(type, from: data)
        } catch {
            fatalError("Couldn't parse \(resource) as \(T.self):\n\(error)")
        }
    }
    
    static func load<T: Decodable>(resource: String) -> T {
        load(resource: resource, type: T.self)
    }
}
// Usage:
let employee1 = JSONLoader.load("employee.json", Employee.self)
let employee2: Employee = JSONLoader.load("employee.json")

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>