从API响应创建xml
本文关键字:xml 创建 响应 API | 更新日期: 2023-09-27 18:01:34
我有以下应用程序,用于发出GET请求,以JSON格式返回数据:
class Program
{
static void Main(string[] args)
{
RunAsync().Wait();
}
static async Task RunAsync()
{
using (var client = new HttpClient())
{
//Send HTTP request
client.BaseAddress = new Uri("httpMyURI");
client.DefaultRequestHeaders.Accept.Clear();
client.DefaultRequestHeaders.Accept.Add(new MediaTypeWithQualityHeaderValue("application/json"));
HttpResponseMessage response = await client.GetAsync("api/people/.....");
if (response.IsSuccessStatusCode)
{
var person = response.Content.ReadAsAsync<IEnumerable<Persons>>().Result;
foreach (var i in person)
{
Console.WriteLine("{0}'t{1}-{2}", i.FirstName, i.LastName, i.Peopleid);
}
}
}
}
}
class Persons
{
public string Peopleid { get; set; }
public string FirstName { get; set; }
public string LastName { get; set; }
}
我如何从我得到的结果创建一个xml文件?我想我首先要转换JSON到xml然后创建一个xml文件?
使用Newtonsoft Json可以很容易地做到这一点。网络图书馆。查看操作中的示例:http://www.newtonsoft.com/json/help/html/convertingjsonandxml.htm
后面的行是
XmlDocument doc = (XmlDocument)JsonConvert.DeserializeXmlNode(json);