Linq-to-XML用于检索所有子节点

Question

I have the following XML

 <?xml version="1.0" encoding="utf-8"?>
 <ErrorServer>
   <ClientIP>
     <AllowAll>false</AllowAll>
     <Client_127_0_0_1>true</Client_127_0_0_1>
   </ClientIP>
   <Users>
     <Admin>
       <Password>passw0r!d</Password>
       <NextError>83</NextError>
       <Active>true</Active>
     </Admin>
     <JimBob>
       <Password>passw0r!d</Password>
       <NextError>83</NextError>
       <Active>true</Active>
     </JimBob>
   </Users>
 </ErrorServer>

Using linq in c# I am trying to get all user names (Admin & JimBob in the example above) using the following code

    List<string> Result = new List<string>();

    XDocument xdoc = XDocument.Load("ErrorServerConfig.xml");

    //Run query
    var lv1s = from lv1 in xdoc.Descendants("ErrorServer")
               select new
               {
                   Children = lv1.Elements("Users")
               };

    //Loop through results
    foreach (var lv1 in lv1s)
    {
        foreach (var lv2 in lv1.Children)
            Result.Add(lv2.Name.ToString());
    }

    return (Result);

Which isn't working as it only returns "Users" in the result.

I am new to linq can anybody tell me the correct way of doing it please?

Answer 1

var result = xdoc.Descendants("Users")
                 .First()
                 .Elements()
                 .Select(e=>e.Name);

Answer 2

Thanks all, I have changed my xml as per Aaron Anodide's sugestion which is what it should have been in the first place (my bad).

Xml now looks like:

<ErrorServer>
  <Users>
    <User>
      <Username>Admin</Username>
      <Password>passw0r!d</Password>
      <NextError>83</NextError>
      <Active>true</Active>
    </User>
  </Users>
</ErrorServer>

and use the following code

1 Select all user names

        XDocument xdoc = XDocument.Load("ErrorServerConfig.xml");

        //Run query
        var result = from e in xdoc.Descendants("Users").Elements()
             select (string)e.Element("Username");

        //Loop through results
        foreach (string user in result)
        {
            Result += String.Format("{0}\r\n", user);
        }

2 get password

    XDocument xdoc = XDocument.Load("ErrorServerConfig.xml");

    //Run query
    var result = (from e in xdoc.Descendants("Users").Elements()
                 where (string)e.Element("Username") == userName
                 select e).Descendants("Password").First().Value;

Both far more elegant than my previous xml code, thanks for your help and pointers.

Answer 3

var res =  XDocument.Load("yourpath")
    .Descendants("Users").Elements()
    .Select(xe => xe.Name.LocalName);

if you are wanting to return a IEnumerable<XName> then use .Name , if you are wanting to reutrn IEnumerable<string> use Name.LocalName . Its just my opinion but in your class i would set the xdoc as a property.

NEW Update

---

this actually got upvoted today (8/7/2014), which prompted me to look at it and critique my own work. I then realized that this is complete crap...

1. As Aaron Anodide commented in the OP question, the Xml architecture was all wrong and that was what was making this such a hard task...

2. Shame on me for enabling crappy code.

3. Shame on me for posting crappy code

The way it should actually be implemented. VVVVV

---

XML:

<?xml version="1.0" encoding="utf-8" ?>
<ErrorServer>
  <ClientIP>
    <AllowAll>false</AllowAll>
    <Address>127.0.0.1</Address>
  </ClientIP>
  <Users>
    <User>
      <Username>Admin</Username>
      <Password>passw0r!d</Password>
      <NextError>83</NextError>
      <Active>true</Active>
    </User>
    <User>
      <Username>JimBob</Username>
      <Password>passw0r!d</Password>
      <NextError>83</NextError>
      <Active>true</Active>
    </User>
  </Users>
</ErrorServer>

Classes:

#region Referencing

using System;
using System.IO;
using System.Linq;
using System.Xml.Serialization;

#endregion

namespace Stack
{
    public class Program
    {
        public Program()
        {
            ErrorServer = ErrorServer.Deserialize( "path" );
        }

        public ErrorServer ErrorServer { get; set; }

        // This way you dont actually have to deal with LINQ and XML.
        // It's just as easy to create a few classes to hold your data, so you can use xml serialization.
        public User GetUserInfoByName( string name )
        {
            return
                ErrorServer.Users.FirstOrDefault(
                    user => user.Username.Equals( name, StringComparison.CurrentCultureIgnoreCase ) );
        }
    }

    [Serializable]
    public class ErrorServer
    {
        public ClientIP ClientIP { get; set; }

        [XmlArrayItem( "User" )]
        public User[] Users { get; set; }

        public static ErrorServer Deserialize( string path )
        {
            using (var stream = new FileStream( path, FileMode.Open ))
                return new XmlSerializer( typeof (ErrorServer) ).Deserialize( stream ) as ErrorServer;
        }
    }

    [Serializable]
    public class ClientIP
    {
        public bool AllowAll { get; set; }

        public string Address { get; set; }
    }

    [Serializable]
    public class User
    {
        public string Username { get; set; }

        public string Password { get; set; }

        public double NextError { get; set; }

        public bool Active { get; set; }
    }
}

So for the love of god, please don't use anything below the line.

---

Update

Sorry it took so long. Here is a little class I put together for you.

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

namespace StackTesting
{
    class Program
    {
        public class User
        {
         public string Username { get; set; }
         public string Pass { get; set; }
         public double Error { get; set; }
         public bool Active { get; set; }

          public User() { }
        }

        Public XDocument xDoc { get; set; }

        static void Main(string[] args)
        {
         xDoc = XDocument.Load(@"C:\Users\Trae\Documents\visual studio 2012\Projects\StackTesting\StackTesting\XMLFile1.xml");
          var user = (User) GetUserInfo("Admin");
        }

        public static User GetUserInfo(string UserName)
        {
          return xDoc.Root.Elements("Users").Elements()
            .Where(xe => xe.Element(XName.Get("Username")).Value == UserName)
            .Select(xe =>
              new User
              {
                Username = xe.Element(XName.Get("Username")).Value,
                Pass = xe.Element(XName.Get("Password")).Value,
                Error = double.Parse(xe.Element(XName.Get("NextError")).Value),
                Active = bool.Parse(xe.Element(XName.Get("Active")).Value)
              }).ToArray()[0];
        }
    }
}

Answer 4

var result = xdoc.Root
                 .Element("Users")
                 .Elements()
                 .Select(x => x.Name);