ios 在 Swift 中将 JSON 字符串转换为 Object 的简单而干净的方法
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Simple and clean way to convert JSON string to Object in Swift
提问by Hasan Nizamani
I have been searching for days to convert a fairly simple JSON string to an object type in Swift but with no avail.
我一直在寻找将相当简单的 JSON 字符串转换为 Swift 中的对象类型的方法，但无济于事。
Here is the code for web service call:
这是 Web 服务调用的代码：
func GetAllBusiness() {
Alamofire.request(.GET, "http://MyWebService/").responseString { (request, response, string, error) in
println(string)
}
}
I have a swift struct Business.swift:
我有一个 swift 结构 Business.swift：
struct Business {
var Id : Int = 0
var Name = ""
var Latitude = ""
var Longitude = ""
var Address = ""
}
Here is my test service deployed:
这是我部署的测试服务：
[
{
"Id": 1,
"Name": "A",
"Latitude": "-35.243256",
"Longitude": "149.110701",
"Address": null
},
{
"Id": 2,
"Name": "B",
"Latitude": "-35.240592",
"Longitude": "149.104843",
"Address": null
}
...
]
It would be a delight if someone guide me through this.
如果有人指导我完成这件事，我会很高兴。
Thanks.
谢谢。
采纳答案by Maxim Shoustin
Here are some tips how to begin with simple example.
以下是一些如何从简单示例开始的提示。
Consider you have following JSON Array String (similar to yours) like:
考虑您有以下 JSON 数组字符串（类似于您的），例如：
var list:Array<Business> = []
// left only 2 fields for demo
struct Business {
var id : Int = 0
var name = ""
}
var jsonStringAsArray = "[\n" +
"{\n" +
"\"id\":72,\n" +
"\"name\":\"Batata Cremosa\",\n" +
"},\n" +
"{\n" +
"\"id\":183,\n" +
"\"name\":\"Caldeirada de Peixes\",\n" +
"},\n" +
"{\n" +
"\"id\":76,\n" +
"\"name\":\"Batata com Cebola e Ervas\",\n" +
"},\n" +
"{\n" +
"\"id\":56,\n" +
"\"name\":\"Arroz de forma\",\n" +
"}]"
// convert String to NSData
var data: NSData = jsonStringAsArray.dataUsingEncoding(NSUTF8StringEncoding)!
var error: NSError?
// convert NSData to 'AnyObject'
let anyObj: AnyObject? = NSJSONSerialization.JSONObjectWithData(data, options: NSJSONReadingOptions(0),
error: &error)
println("Error: \(error)")
// convert 'AnyObject' to Array<Business>
list = self.parseJson(anyObj!)
//===============
func parseJson(anyObj:AnyObject) -> Array<Business>{
var list:Array<Business> = []
if anyObj is Array<AnyObject> {
var b:Business = Business()
for json in anyObj as Array<AnyObject>{
b.name = (json["name"] as AnyObject? as? String) ?? "" // to get rid of null
b.id = (json["id"] as AnyObject? as? Int) ?? 0
list.append(b)
}// for
} // if
return list
}//func
[EDIT]
[编辑]
To get rid of null changed to:
去掉 null 改为：
b.name = (json["name"] as AnyObject? as? String) ?? ""
b.id = (json["id"] as AnyObject? as? Int) ?? 0
See also Reference of Coalescing Operator(aka ??)
另请参阅Coalescing Operator（又名??）的参考
Hope it will help you to sort things out,
希望能帮到你解决问题
回答by FlowUI. SimpleUITesting.com
for swift 3/4
快速 3/4
extension String {
func toJSON() -> Any? {
guard let data = self.data(using: .utf8, allowLossyConversion: false) else { return nil }
return try? JSONSerialization.jsonObject(with: data, options: .mutableContainers)
}
}
Example Usage:
示例用法：
let dict = myString.toJSON() as? [String:AnyObject] // can be any type here
回答by PassKit
As simple String extension should suffice:
简单的 String 扩展就足够了：
extension String {
var parseJSONString: AnyObject? {
let data = self.dataUsingEncoding(NSUTF8StringEncoding, allowLossyConversion: false)
if let jsonData = data {
// Will return an object or nil if JSON decoding fails
return NSJSONSerialization.JSONObjectWithData(jsonData, options: NSJSONReadingOptions.MutableContainers, error: nil)
} else {
// Lossless conversion of the string was not possible
return nil
}
}
}
Then:
然后：
var jsonString = "[\n" +
"{\n" +
"\"id\":72,\n" +
"\"name\":\"Batata Cremosa\",\n" +
"},\n" +
"{\n" +
"\"id\":183,\n" +
"\"name\":\"Caldeirada de Peixes\",\n" +
"},\n" +
"{\n" +
"\"id\":76,\n" +
"\"name\":\"Batata com Cebola e Ervas\",\n" +
"},\n" +
"{\n" +
"\"id\":56,\n" +
"\"name\":\"Arroz de forma\",\n" +
"}]"
let json: AnyObject? = jsonString.parseJSONString
println("Parsed JSON: \(json!)")
println("json[3]: \(json![3])")
/* Output:
Parsed JSON: (
{
id = 72;
name = "Batata Cremosa";
},
{
id = 183;
name = "Caldeirada de Peixes";
},
{
id = 76;
name = "Batata com Cebola e Ervas";
},
{
id = 56;
name = "Arroz de forma";
}
)
json[3]: {
id = 56;
name = "Arroz de forma";
}
*/
回答by Noyer282
Swift 4parses JSON much more elegantly. Just adopt the codable protocol for your structure as per this simplified example:
Swift 4更优雅地解析 JSON。按照这个简化的示例，只需为您的结构采用可编码的协议：
struct Business: Codable {
let id: Int
let name: String
}
To parse the JSON array, you tell the decoder what the objects of the data array are
要解析 JSON 数组，您需要告诉解码器数据数组的对象是什么
let parsedData = decoder.decode([Business].self, from: data)
Here's a full working example:
这是一个完整的工作示例：
import Foundation
struct Business: Codable {
let id: Int
let name: String
}
// Generating the example JSON data:
let originalObjects = [Business(id: 0, name: "A"), Business(id: 1, name: "B")]
let encoder = JSONEncoder()
let data = try! encoder.encode(originalObjects)
// Parsing the data:
let decoder = JSONDecoder()
let parsedData = try! decoder.decode([Business].self, from: data)
For more background, check out this excellent guide.
有关更多背景信息，请查看此优秀指南。
回答by swiftBoy
For Swift 4
对于Swift 4
I used @Passkit's logic but i had to update as per Swift 4
我使用了 @ Passkit的逻辑，但我必须按照 Swift 4 进行更新
Step.1 Created extension for String Class
Step.1为字符串类创建扩展
import UIKit
extension String
{
var parseJSONString: AnyObject?
{
let data = self.data(using: String.Encoding.utf8, allowLossyConversion: false)
if let jsonData = data
{
// Will return an object or nil if JSON decoding fails
do
{
let message = try JSONSerialization.jsonObject(with: jsonData, options:.mutableContainers)
if let jsonResult = message as? NSMutableArray
{
print(jsonResult)
return jsonResult //Will return the json array output
}
else
{
return nil
}
}
catch let error as NSError
{
print("An error occurred: \(error)")
return nil
}
}
else
{
// Lossless conversion of the string was not possible
return nil
}
}
}
Step.2 This is how I used in my view controller
Step.2这就是我在视图控制器中的使用方式
var jsonString = "[\n" +
"{\n" +
"\"id\":72,\n" +
"\"name\":\"Batata Cremosa\",\n" +
"},\n" +
"{\n" +
"\"id\":183,\n" +
"\"name\":\"Caldeirada de Peixes\",\n" +
"},\n" +
"{\n" +
"\"id\":76,\n" +
"\"name\":\"Batata com Cebola e Ervas\",\n" +
"},\n" +
"{\n" +
"\"id\":56,\n" +
"\"name\":\"Arroz de forma\",\n" +
"}]"
//Convert jsonString to jsonArray
let json: AnyObject? = jsonString.parseJSONString
print("Parsed JSON: \(json!)")
print("json[2]: \(json![2])")
All credit goes to original user, I just updated for latest swift version
所有功劳归于原始用户，我刚刚更新了最新的 swift 版本
回答by isair
I wrote a library which makes working with json data and deserialization a breeze in Swift. You can get it here: https://github.com/isair/JSONHelper
我编写了一个库，它使在 Swift 中处理 json 数据和反序列化变得轻而易举。你可以在这里得到它：https: //github.com/isair/JSONHelper
Edit: I updated my library, you can now do it with just this:
编辑：我更新了我的图书馆，你现在可以这样做：
class Business: Deserializable {
var id: Int?
var name = "N/A" // This one has a default value.
required init(data: [String: AnyObject]) {
id <-- data["id"]
name <-- data["name"]
}
}
var businesses: [Business]()
Alamofire.request(.GET, "http://MyWebService/").responseString { (request, response, string, error) in
businesses <-- string
}
Old Answer:
旧答案：
First, instead of using .responseString, use .response to get a response object. Then change your code to:
首先，不要使用 .responseString，而是使用 .response 来获取响应对象。然后将您的代码更改为：
func getAllBusinesses() {
Alamofire.request(.GET, "http://MyWebService/").response { (request, response, data, error) in
var businesses: [Business]?
businesses <-- data
if businesses == nil {
// Data was not structured as expected and deserialization failed, do something.
} else {
// Do something with your businesses array.
}
}
}
And you need to make a Business class like this:
你需要像这样制作一个商业类：
class Business: Deserializable {
var id: Int?
var name = "N/A" // This one has a default value.
required init(data: [String: AnyObject]) {
id <-- data["id"]
name <-- data["name"]
}
}
You can find the full documentation on my GitHub repo. Have fun!
您可以在我的 GitHub 存储库上找到完整的文档。玩得开心！
回答by Timeless
For iOS 10& Swift 3, using Alamofire& Gloss:
对于iOS 10& Swift 3，使用Alamofire& Gloss：
Alamofire.request("http://localhost:8080/category/en").responseJSON { response in
if let data = response.data {
if let categories = [Category].from(data: response.data) {
self.categories = categories
self.categoryCollectionView.reloadData()
} else {
print("Casting error")
}
} else {
print("Data is null")
}
}
and here is the Category class
这是类别类
import Gloss
struct Category: Decodable {
let categoryId: Int?
let name: String?
let image: String?
init?(json: JSON) {
self.categoryId = "categoryId" <~~ json
self.name = "name" <~~ json
self.image = "image" <~~ json
}
}
IMO, this is by far the most elegant solution.
IMO，这是迄今为止最优雅的解决方案。
回答by Danny Narváez
For Swift 4, i wrote this extension using the Codableprotocol:
对于Swift 4，我使用Codable协议编写了这个扩展：
struct Business: Codable {
var id: Int
var name: String
}
extension String {
func parse<D>(to type: D.Type) -> D? where D: Decodable {
let data: Data = self.data(using: .utf8)!
let decoder = JSONDecoder()
do {
let _object = try decoder.decode(type, from: data)
return _object
} catch {
return nil
}
}
}
var jsonString = "[\n" +
"{\n" +
"\"id\":72,\n" +
"\"name\":\"Batata Cremosa\",\n" +
"},\n" +
"{\n" +
"\"id\":183,\n" +
"\"name\":\"Caldeirada de Peixes\",\n" +
"},\n" +
"{\n" +
"\"id\":76,\n" +
"\"name\":\"Batata com Cebola e Ervas\",\n" +
"},\n" +
"{\n" +
"\"id\":56,\n" +
"\"name\":\"Arroz de forma\",\n" +
"}]"
let businesses = jsonString.parse(to: [Business].self)
回答by MR_22
I like RDC's response, but why limit the JSON returned to have only arrays at the top level? I needed to allow a dictionary at the top level, so I modified it thus:
我喜欢 RDC 的响应，但为什么将返回的 JSON 限制为只有顶级数组？我需要在顶层允许字典，所以我修改了它：
extension String
{
var parseJSONString: AnyObject?
{
let data = self.dataUsingEncoding(NSUTF8StringEncoding, allowLossyConversion: false)
if let jsonData = data
{
// Will return an object or nil if JSON decoding fails
do
{
let message = try NSJSONSerialization.JSONObjectWithData(jsonData, options:.MutableContainers)
if let jsonResult = message as? NSMutableArray {
return jsonResult //Will return the json array output
} else if let jsonResult = message as? NSMutableDictionary {
return jsonResult //Will return the json dictionary output
} else {
return nil
}
}
catch let error as NSError
{
print("An error occurred: \(error)")
return nil
}
}
else
{
// Lossless conversion of the string was not possible
return nil
}
}
回答by Munish Kapoor
let jsonString = "{\"id\":123,\"Name\":\"Munish\"}"
Convert String to NSData
将字符串转换为 NSData
var data: NSData =jsonString.dataUsingEncoding(NSUTF8StringEncoding)!
var error: NSError?
Convert NSData to AnyObject
将 NSData 转换为 AnyObject
var jsonObject: AnyObject? = NSJSONSerialization.JSONObjectWithData(data, options: NSJSONReadingOptions.allZeros, error: &error)
println("Error: \(error)")
let id = (jsonObject as! NSDictionary)["id"] as! Int
let name = (jsonObject as! NSDictionary)["name"] as! String
println("Id: \(id)")
println("Name: \(name)")
