C＃：在M解决方案中放入许多Json供将来使用

Question

所以我有this Json列表。

我想将此列表放入我的解决方案中，并不时询问特定的国家/地区，并获取其所有Json及其所有参数（也不是所有必需的参数）

有什么想法吗？

Answer 1

这是一个非常简单的对象。最简单的方法是将其转换为C＃对象，然后将其包含在解决方案中，而不是将其存储为原始JSON对象并进行解析。

Answer 2

As @ADyson said in his comment there is lot of documentation out there to help you educate and implement the scenario you mentioned. 

using Newtonsoft.Json;
using Newtonsoft.Json.Linq;

//Creating a Class object with all the properties of json string
public class CountriesInfo
{
    public string type { get; set; }
    public string countryName { get; set; }
    public string countryCode { get; set; }
    public string mcc { get; set; }
    public string mnc { get; set; }
    public string brand { get; set; }
    public string @operator { get; set; }
    public string status { get; set; }
    public string bands { get; set; }
    public string notes { get; set; }
}

  static void Main(string[] args)
    {
        var strCountriesList = File.ReadAllText(@"YourJSONFilePath");

        JArray countriesInfoListArray = JArray.Parse(strCountriesList);

        List<CountriesInfo> countriesInfoList = countriesInfoListArray.Select(p => new CountriesInfo()
        {
            type = (string) p["type"],
            countryName = (string) p["countryName"],
            countryCode = (string) p["countryCode"],
            mcc = (string) p["mcc"],
            mnc = (string) p["mnc"],
            brand = (string) p["brand"],
            @operator = (string) p["@operator"],
            status = (string) p["status"],
            bands = (string) p["bands"],
            notes = (string) p["notes"]

        }).ToList();

       // Filters the countries based on the countryName (since the country name has null I am checking for null in countryName
        var filteredCountries =  countriesInfoList.Where(p => p.countryName != null && p.countryName.Equals("CountryYouWantToFilter"));

        foreach (var countryInfo in filteredCountries)
        {
           //do something with the countryInfo as shown below
            var strOperator = countryInfo.@operator;
            var status = countryInfo.status; 
        }