Question

在浏览器中，我的网址以完美的JSON格式给出结果，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对json数据进行解码

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()

请帮助我解决这个问题。

Answer 1

首先，您必须创建一个Decodable结构并将其放在viewController类之前：

 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struct数组var：

var myVar = [YourArrray]()

现在您可以继续解码json：

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，您只需调用fetchJsonObject（）函数即可完成希望有帮助：）

我的json响应的格式与我预期的不同

1 个答案: