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

这在XCode 8.3.3中是可行的

func fetchPersons(){

    if let pathURL = Bundle.main.url(forResource: "Person", withExtension: "json"){

        do {

            let jsonData = try Data(contentsOf: pathURL, options: .mappedIfSafe)

            let jsonResult = try JSONSerialization.jsonObject(with: jsonData, options: .mutableContainers) as! [String: Any]
            if let persons = jsonResult["person"] as? [Any]{

                print(persons)
            }

        }catch(let error){
            print (error.localizedDescription)
        }
    }
}

其他回答

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

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")
*/

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
      }
}

Swift 4 JSON类与可解码-为那些喜欢类

定义类如下:

class People: Decodable {
  var person: [Person]?

  init(fileName : String){
    // url, data and jsonData should not be nil
    guard let url = Bundle.main.url(forResource: fileName, withExtension: "json") else { return }
    guard let data = try? Data(contentsOf: url) else { return }
    guard let jsonData = try? JSONDecoder().decode(People.self, from: data) else { return }

    // assigns the value to [person]
    person = jsonData.person
  }
}

class Person : Decodable {
  var name: String
  var age: String
  var employed: String
}

用法,非常抽象:

let people = People(fileName: "people")
let personArray = people.person

这允许People类和Person类的方法,如果需要,变量(属性)和方法也可以标记为private。

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。