在Swift中将JSON字符串转换为对象的简单明了的方法
在Swift中将JSON字符串转换为对象的简单明了的方法
提问于 2014-09-02 18:23:17
几天来，我一直在寻找在Swift中将一个相当简单的JSON字符串转换为对象类型，但没有结果。
以下是web服务调用的代码：
func GetAllBusiness() {
Alamofire.request(.GET, "http://MyWebService/").responseString { (request, response, string, error) in
println(string)
}
}我有一个快速结构Business.swift：
struct Business {
var Id : Int = 0
var Name = ""
var Latitude = ""
var Longitude = ""
var Address = ""
}下面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
}
...
]如果有人能指导我完成这件事，那将是一件令人愉快的事情。
谢谢。
回答 9
Stack Overflow用户
发布于 2017-02-22 07:46:44
swift 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)
}
}示例用法：
let dict = myString.toJSON() as? [String:AnyObject] // can be any type hereStack Overflow用户
发布于 2014-12-03 18:28:18
因为简单的字符串扩展应该足够了：
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
}
}
}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";
}
*/Stack Overflow用户
发布于 2017-03-18 22:38:06
对于iOS 10和Swift 3，使用Alamofire和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")
}
}这是Category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
}
}到目前为止，这是最优雅的解决方案。
页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/25621120
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