如何使用 ZD7EFA19FBE7D3972FD5ADB6024223D74 将 JSON object 解析为 class

Question

My Json Object

{"returnStatus":{"code":"0200","status":"SUCCESS","message":"The operation was successful.","requestTime":"20210407141112083","responseTime":"20210407141112256"},"allCities":[{"cityID":2622,"cityName":" UNKNOWN","cityCode":"AAA","area":"KHI"}

I want to parse into class.How? How to get cityID,CityName,cityCode,area etc..

I'm trying this code

allCities Cities = JsonConvert.DeserializeObject<dynamic>(response);

Answer 1

You can use like Ryan say before an online tool to create classes from your json. But i prefer to use the integrated "Paste Special" Function from Visual Studio.

Answer 2

You can use Json2CSharp to create classes to deserilalize your object to. You can rename them accordingly if you wish.

// Root myDeserializedClass = JsonConvert.DeserializeObject<Root>(myJsonResponse); 
public class ReturnStatus
{
    public string code { get; set; }
    public string status { get; set; }
    public string message { get; set; }
    public string requestTime { get; set; }
    public string responseTime { get; set; }
}

public class AllCity
{
    public int cityID { get; set; }
    public string cityName { get; set; }
    public string cityCode { get; set; }
    public string area { get; set; }
}

public class Root
{
    public ReturnStatus returnStatus { get; set; }
    public List<AllCity> allCities { get; set; }
}

You can then use myDeserializedClass.allCities to access the list of cities.