从SHAREPOINT结果读取XML值

Question

我是否可以询问如何在foreach循环或for循环中从SHAREPOINT结果读取XML值？

这里是我的源代码，我想阅读d：、d:Description和d:W循/d:Url的值

XmlNamespaceManager xmlnspm = new XmlNamespaceManager(new NameTable());
Uri sharepointUrl = new Uri("SHAREPOINT URL);

xmlnspm.AddNamespace("atom", "http://www.w3.org/2005/Atom");
xmlnspm.AddNamespace("d", "http://schemas.microsoft.com/ado/2007/08/dataservices");
xmlnspm.AddNamespace("m", "http://schemas.microsoft.com/ado/2007/08/dataservices/metadata");

NetworkCredential cred = new System.Net.NetworkCredential("USERNAME", "PASSWORD", "DOMAIN");

HttpWebRequest listRequest = (HttpWebRequest)HttpWebRequest.Create(sharepointUrl.ToString() + "_api/lists/getByTitle('" + "LIST NAME" + "')/items");
listRequest.Method = "GET";
listRequest.Accept = "application/atom+xml";
listRequest.ContentType = "application/atom+xml;type=entry";

listRequest.Credentials = cred;
HttpWebResponse listResponse = (HttpWebResponse)listRequest.GetResponse();
StreamReader listReader = new StreamReader(listResponse.GetResponseStream());
XmlDocument listXml = new XmlDocument();

listXml.LoadXml(listReader.ReadToEnd());

我的示例XML文件：

<?xml version="1.0" encoding="UTF-8"?>
<feed xmlns="http://www.w3.org/2005/Atom" xmlns:d="http://schemas.microsoft.com/ado/2007/08/dataservices" xmlns:georss="http://www.georss.org/georss" xmlns:gml="http://www.opengis.net/gml" xmlns:m="http://schemas.microsoft.com/ado/2007/08/dataservices/metadata" xml:base="https://mvponduty.sharepoint.com.com/sites/sg/daw/_api/">
   <id>ce656942-44db-472b-93cd-0caeb82f8ffb</id>
   <title />
   <updated>2019-12-03T01:52:11Z</updated>
   <entry m:etag="&quot;8&quot;">
      <content type="application/xml">
         <m:properties>
            <d:Title>User guide - Bloomberg for advanced users</d:Title>
            <d:Follow m:type="SP.FieldUrlValue">
               <d:Description>Follow</d:Description>
               <d:Url>https://mvponduty.sharepoint.com/sites/sg/daw/_layouts/SubNew.aspx?List=%7B3EE60A7B%2D88C4%2D44E4%2D81AC%2DFEC9D91764F5%7D&amp;ID=113&amp;Source=http%3A%2F%2Fportal%2Easpac%2Ekworld%2Ekpmg%2Ecom%2Fsg%2Fdw%2FPages%2Fdefault%2Easpx</d:Url>
            </d:Follow>
         </m:properties>
      </content>
   </entry>
   <entry m:etag="&quot;8&quot;">
      <content type="application/xml">
         <m:properties>
            <d:Title>User guide - Bloomberg for advanced users</d:Title>
            <d:Follow m:type="SP.FieldUrlValue">
               <d:Description>Follow</d:Description>
               <d:Url>https://mvponduty.sharepoint.com/sites/sg/daw/_layouts/SubNew.aspx?List=%7B3EE60A7B%2D88C4%2D44E4%2D81AC%2DFEC9D91764F5%7D&amp;ID=113&amp;Source=http%3A%2F%2Fportal%2Easpac%2Ekworld%2Ekpmg%2Ecom%2Fsg%2Fdw%2FPages%2Fdefault%2Easpx</d:Url>
            </d:Follow>
         </m:properties>
      </content>
   </entry>
</feed>

Answer 1

我建议您使用JSON访问REST，我们可以更容易地获得结果。

示例代码

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


namespace REST  
{  
    class Program  
    {  
        static void Main(string[] args)  
        {  
            HttpWebRequest endpointRequest = (HttpWebRequest)HttpWebRequest.Create("Site URL/_api/web/lists/getByTitle(List Name')/items");  

            endpointRequest.Method = "GET";  
            endpointRequest.Accept = "application/json;odata=verbose";  
            NetworkCredential cred = new System.Net.NetworkCredential("username", "password", "domain");  
            endpointRequest.Credentials = cred;  
            HttpWebResponse endpointResponse = (HttpWebResponse)endpointRequest.GetResponse();  
            try  
            {  
                WebResponse webResponse = endpointRequest.GetResponse();  
                Stream webStream = webResponse.GetResponseStream();  
                StreamReader responseReader = new StreamReader(webStream);  
                string response = responseReader.ReadToEnd();  
                JObject jobj = JObject.Parse(response);  
                JArray jarr = (JArray)jobj["d"]["results"];  
                foreach (JObject j in jarr)  
                {  
                    Console.WriteLine(j["Title"]+" "+j["Body"]);  
                }  

                responseReader.Close();  
                Console.ReadLine();  


            }  
            catch (Exception e)  
            {  
                Console.Out.WriteLine(e.Message); Console.ReadLine();  
            }       
        }  
    }  
} 

Answer 2

在尝试了几种方法之后，我找到了解决办法。

这里有我的示例源代码。

    // Get and display all the title and Description
                XmlElement root = listXml.DocumentElement;
                XmlNodeList elemList = root.GetElementsByTagName("content");
                XmlNodeList elemList_title = root.GetElementsByTagName("d:Title");
                XmlNodeList elemList_desc = root.GetElementsByTagName("d:Description");



                for (int i = 0; i < elemList.Count; i++)
                {

                    string title = elemList_title[i].InnerText;
                    string description = elemList_desc[i].InnerText;
            Console.WriteLine(title+description);

                }