無法使用的XPathNavigator

Question

List<string> list = new List<string>(); 
foreach (XPathNavigator node in nav.Select("configuration/company/work/worktime")) 
      { 
       string day = getAttribute(node, "day"); 
       string time = getAttribute(node, "time"); 
       string worktype = ?? // how to get worktype attribute valuefrom parent node 
       list.Add(day,time,worktype); // add to list 
      } 

</configuration> 
     <company> 
     <work worktype="homeWork"> 
      <worktime day="30" time="10:28"></worktime> 
      <worktime day="25" time="10:50"></worktime> 
     </work> 
     <work worktype="officeWork"> 
      <worktime day="12" time="09:28"></worktime> 
      <worktime day="15" time="12:28"></worktime> 
     </work> 
     </company> 
    </configuration> 


need output as : 
list[0] = homeWork,30,10:28 
list[1] = homeWork,25,10:50 
list[2] = officeWork,12,09:28 
list[3] = officeWork,15,12:28 

我想從XML名單，但未能像上面給出的（使用XPath導航器獲取輸出，我怎麼能訪問父節點得到worktype屬性從XML獲取列表，以及其他剩餘內部節點屬性？

Answer 1

我建議使用LINQ to XML了XPath的，但如果你必須使用XPathNavigator那麼你需要遍歷每個work元素，然後它的每一個worktime子元素。這樣，您就可以從父上下文中使用worktype：

foreach (XPathNavigator work in nav.Select("configuration/company/work")) 
{ 
    var workType = work.GetAttribute("worktype", string.Empty); 

    foreach (XPathNavigator worktime in work.Select("worktime")) 
    { 
     var day = worktime.GetAttribute("day", string.Empty); 
     var time = worktime.GetAttribute("time", string.Empty); 

     list.Add($"{workType}, {day}, {time}"); 
    } 
} 

了工作演示見this fiddle。

Answer 2

使用嵌套循環。最初configuration/company/work檢索工作節點，檢索一個變量worktype屬性和存儲。然後，通過孩子worktype節點環和一個字符串添加到列表中的每一個

Answer 3

使用Net Library增強版xml（linq xml）

using System; 
using System.Collections.Generic; 
using System.Linq; 
using System.Text; 
using System.Xml; 
using System.Xml.Linq; 

namespace ConsoleApplication1 
{ 
    class Program 
    { 
     const string FILENAME = @"c:\temp\test.xml"; 
     static void Main(string[] args) 
     { 
      XDocument doc = XDocument.Load(FILENAME); 
      var results = doc.Descendants("work").Select(x => new { 
       worktype = (string)x.Attribute("worktype"), 
       worktime = x.Elements("worktime").Select(y => new { 
        day = (int)y.Attribute("day"), 
        time = (DateTime)y.Attribute("time") 
       }).ToList() 
      }).ToList(); 
     } 
    } 
}