Linq-to-XML,用于检索所有子节点
本文关键字:子节点 检索 用于 Linq-to-XML | 更新日期: 2023-09-27 18:28:56
我有以下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>
使用c#中的linq,我试图使用以下代码获取所有用户名(上面示例中的Admin和JimBob)
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);
这不起作用,因为它只在结果中返回"Users"。
我是林克的新手,有人能告诉我正确的做法吗?
var result = xdoc.Descendants("Users")
.First()
.Elements()
.Select(e=>e.Name);
谢谢大家,我已经按照Aaron Anode的建议更改了我的xml,这是它最初应该是的(我的错)。
Xml现在看起来像:
<ErrorServer>
<Users>
<User>
<Username>Admin</Username>
<Password>passw0r!d</Password>
<NextError>83</NextError>
<Active>true</Active>
</User>
</Users>
</ErrorServer>
并使用以下代码
1选择所有用户名
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获取密码
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;
两者都比我以前的xml代码优雅得多,感谢您的帮助和指导。
var res = XDocument.Load("yourpath")
.Descendants("Users").Elements()
.Select(xe => xe.Name.LocalName);
如果您想返回IEnumerable<XName>
,则使用.Name
;如果您想重新使用IEnumerable<string>
,请使用Name.LocalName
。这只是我的意见,但在你的课上,我会把xdoc设置为一个属性。
最新更新
事实上,这在今天(2014年8月7日)得到了支持,这促使我审视它并批评我自己的作品。然后我意识到这完全是垃圾。。。
正如Aaron Anode在OP问题中所评论的那样,Xml体系结构完全错误,这就是为什么这是一项艰巨的任务。。。
我为启用糟糕的代码而感到羞愧。
为我张贴糟糕的代码感到羞耻
它的实际实施方式。vvvv
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>
类别:
#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; }
}
}
所以,为了上帝的爱,请不要使用任何低于底线的东西
更新
抱歉花了这么长时间。这是我为你整理的一个小课堂。
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];
}
}
}
var result = xdoc.Root
.Element("Users")
.Elements()
.Select(x => x.Name);