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。
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)")
}
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。
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
}
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")
*/
