c# - 从页面读取 XML 响应

Question

我正在使用 C# 和 ASP.net 对网页执行 POST。如何读取 XML 响应以了解我的提交是否有误或成功？

这是我尝试过的方法，但它只会返回成功/失败消息，不会显示从页面返回的实际 xml。

private void Perform()
{
    this.lblResult.Text = string.Empty;
    Dictionary<string, string> dictFormValues = new Dictionary<string, string>();
    string connectionString = null;
    SqlConnection cnn;
    SqlCommand cmd;
    StringBuilder sql = new StringBuilder();
    SqlDataReader reader;
    string email = string.Empty;
    connectionString = "Data Source=server;Initial Catalog=db;User ID=;Password=";
    sql.Append("select TOP 1 maexst ");
    sql.Append("from redbone.redlight.dbo.maxima ");

    cnn = new SqlConnection(connectionString);
    try
    {
        cnn.Open();
        cmd = new SqlCommand(sql.ToString(), cnn);
        reader = cmd.ExecuteReader();
        while (reader.Read()) { dictFormValues.Add("maexst", reader.GetValue(0).ToString()); }
        reader.Close();
        cmd.Dispose();
        cnn.Close();
    }
    catch (Exception ex) { Response.Write(ex.Message.ToString()); }
    string strIpAddress = System.Web.HttpContext.Current.Request.UserHostAddress;
    string strPageTitle = this.Title;
    string strPageURL = System.Web.HttpContext.Current.Request.Url.AbsoluteUri;
    string strError = "";
    bool blnRet = false;
    blnRet = Post(dictFormValues, strPageTitle, strPageURL, ref strError);
    if (blnRet == true)
    {
        this.lblResult.Text = "It was good!";
    }
    else { this.lblResult.Text = strError + ": Error Occured"; }
}

public bool blnRet(Dictionary<string, string> dictFormValues, string strPageTitle, string strPageURL, ref string strMessage)
{
    string strEndpointURL = string.Format("http://testtest12test123.aspx");

    System.Web.Script.Serialization.JavaScriptSerializer json = new System.Web.Script.Serialization.JavaScriptSerializer();
    string strPostData = "";
    foreach (var d in dictFormValues) { strPostData += d.Key + "=" + Server.UrlEncode(d.Value) + "&"; }
    strPostData += "hs_context=";
    System.Net.HttpWebRequest r = (System.Net.HttpWebRequest)System.Net.WebRequest.Create(strEndpointURL);
    r.Method = "POST";
    r.Accept = "application/json";
    r.ContentType = "application/x-www-form-urlencoded";
    r.ContentLength = strPostData.Length;
    r.KeepAlive = false;
    using (System.IO.StreamWriter sw = new System.IO.StreamWriter(r.GetRequestStream()))
    {
        try { sw.Write(strPostData); }
        catch (Exception ex)
        {
            strMessage = ex.Message;
            return false;
        }
    }
    return true; 
}

编辑

成功响应

<?xml version="1.0" encoding="utf-8" ?>
<result>
   <success>1</success>
   <postid>12345</postid>
   <errors/>
</result>

响应失败

<?xml version="1.0" encoding="utf-8" ?>
<result>
   <success>0</success>
   <postid/>
   <errors>
      <error>Error Listed Here</error>
      <error>Error 2 Listed Here</error>
      <error>Error 3 Listed Here</error>
   </errors>
</result>

Answer 1

尝试:

var request = WebRequest.Create("http://some.website/") as HttpWebRequest; 
var response = request.GetResponse();

Stream receiveStream = response.GetResponseStream();
StreamReader readStream = new StreamReader(receiveStream, Encoding.UTF8);

var result =  readStream.ReadToEnd();

这会将页面的内容放入result

下一步做什么取决于实际响应是什么......从那里你可以使用:

• XmlDocument.LoadXml
• XDocument.Parse
• XElement.Parse

或者可能类似于 HTML Agility Pack会让你解析响应。

---

使用 XElement 的示例

using System.Xml.Linq;
using System.Linq;
using System.Xml;

var xml = System.Xml.Linq.XElement.Parse(result);
if (xml.Elements("success").FirstOrDefault().Value == "1")
{
   // Process Success
   Console.WriteLine("All Worked!");
}
else
{
   var errors = xml.Elements("errors");
   foreach (var error in errors.Elements("error"))
   {
     // read error messages
     Console.WriteLine(error.Value);
   }
}

( Runnable Fiddle )

可能有一种更简单的方法来解析 XElement，但这应该能让您有所了解。