我的 json 响应格式与我预期的不同

Question

在浏览器中，我的网址给出了完美的JSON格式的结果，如下所示

"articles": [
{
"source": {
"id": "the-times-of-india",
"name": "The Times of India"
},
"author": "Times Of India",

但是在 Xcode 输出中，我得到的响应如下。如何将此响应转换为完美的 json 格式

{
articles =     (
{
author = "Times Of India";
content = "Hyderabad: Senior Police officials arrive at the site of the encounter. All four accused in the rape 
description = "India News: All four accused in the rape and murder of woman veterinarian in Telangana have been killed in an encounter with the police. Cops claimed they tried t";
publishedAt = "2019-12-06T04:15:00Z";
source =             {
name = "The Times of India";
};
},

我正在使用以下代码来解码 json 数据

let task = URLSession.shared.dataTask(with: url) { (data, response, error) in
guard let dataResponse = data, error == nil else {
print(error?.localizedDescription ?? "Response Error")
return
}
do{
//here dataResponse received from a network request
let jsonResponse = try JSONSerialization.jsonObject(with: dataResponse, options: [])
print(jsonResponse) //Response result
} catch let parsingError {
print("Error", parsingError)
}
}
task.resume()

请帮我解决这个问题。

Answer 1

首先，您必须创建一个可解码结构并将其放在viewController class之前：

struct YourArrray: Decodable {
let author: String?
let content: String?
let location: String?
let description : String?
let publishedAt : String?
let name: String?
}

声明您的网址：

let jsonUrlString = "https://yourUrljson"

之后创建您的结构数组 var：

var myVar = [YourArrray]()

现在，您可以对 JSON 进行解码：

fileprivate func fetchJsonObject() {
guard let url = URL(string: jsonUrlString) else { return }
URLSession.shared.dataTask(with: url) { (data, respons, err) in
guard let data = data else { return }
do {
let jsonResponse = try JSONDecoder().decode([myVar].self, from: data)
print(jsonResponse)
} catch let jsonErr {
print("Error serializing:", jsonErr)
}
}.resume()
}

现在你可以简单地调用函数fetchJsonObject((就完成了 希望这对:)有所帮助