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。
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")
*/
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。
在清理和抛光我的代码之后，我来到了这两个函数，你可以添加到你的项目中，并使用它们非常整洁和快速地从json文件读取数据，并将数据转换为你想要的任何类型!
public func readDataRepresentationFromFile(resource: String, type: String) -> Data? {
let filePath = Bundle.main.path(forResource: resource, ofType: type)
if let path = filePath {
let result = FileManager.default.contents(atPath: path)
return result
}
return nil
}
然后在这个函数的帮助下，你可以将你的数据转换为任何你想要的类型:
public func getObject<T: Codable>(of type: T.Type, from file: String) -> T? {
guard let data = readDataRepresentationFromFile(resource: file, type: "json") else {
return nil
}
if let object = try? JSONDecoder().decode(type, from: data) {
return object
}
return nil
}
此代码的应用示例:
在你的代码中调用这个函数，给它你的json文件的名字，这就是你所需要的!
func getInputDataFromSomeJson(jsonFileName: String) -> YourReqiuredOutputType? {
return getObject(of: YourReqiuredOutputType.self, from: jsonFileName)
}
