C# 如何使用HttpClient读取XML响应？

显然，

HttpClient

是新推荐的HTTP请求生成方法，因此我尝试使用它向Delicious API发出请求，该API返回XML响应。以下是我得到的：

internal class Program
{
    private static void Main(string[] args)
    {
        var credentials = new NetworkCredential("username", "password");
        var handler = new HttpClientHandler { Credentials = credentials};
        var client = new HttpClient(handler);

        var suggest = new Uri("https://api.del.icio.us/v1/posts/suggest");

        var suggestions =
            client.GetAsync(suggest.AddQueryParams("url", "https://yahoo.com"))
                .ContinueWith(t => t.Result.Content.ReadAsAsync<DeliciousSuggest>())
                .Unwrap()
                .Result;



        Console.ReadLine();
    }
}

public class DeliciousSuggest
{
    public string[] Popular { get; set; }
    public string[] Recommended { get; set; }
    public string[] Network { get; set; }
}

---

如何将其解析为可用格式？

“如何将其解析为可用格式？”

    [XmlRoot("suggest")]
public class DeliciousSuggest {
    [XmlElement("popular")]
    public string[] Popular { get; set; }

    [XmlElement("recommended")]
    public string[] Recommended { get; set; }

    [XmlElement("network")]
    public string[] Network { get; set; }
}

并使用XmlSerializer进行反序列化

---

您应该将del.icio.us的响应作为字符串读回，然后可以按如下方式反序列化它：

var s = "this is the response from del"; 
var buffer = Encoding.UTF8.GetBytes(s); 
using (var stream = new MemoryStream(buffer)) { 
    var serializer = new XmlSerializer(typeof(DeliciousSuggest)); 
    var deliciousSuggest = (DeliciousSuggest)serializer.Deserialize(stream); 
    //then do whatever you want
}

---

您可以像下面这样将头设置为请求xml。您是说我应该将响应作为字符串读取，并使用

XmlSerializer

作为单独的步骤对其进行解析吗？var s=“这是来自del的响应”；var buffer=Encoding.UTF8.GetBytes（s）；使用（var stream=new MemoryStream（buffer））{var serializer=new XmlSerializer（typeof（DeliciousSuggest））；var DeliciousSuggest=（DeliciousSuggest）serializer.Deserialize（stream）；//然后做任何你想做的事}我认为是肯定的。我用你提供的代码编辑了你的答案，使它更清晰。谢谢你的帮助！

var s = "this is the response from del"; 
var buffer = Encoding.UTF8.GetBytes(s); 
using (var stream = new MemoryStream(buffer)) { 
    var serializer = new XmlSerializer(typeof(DeliciousSuggest)); 
    var deliciousSuggest = (DeliciousSuggest)serializer.Deserialize(stream); 
    //then do whatever you want
}