从根节点获取属性值 [英] Get attribute value from a root node

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本文介绍了从根节点获取属性值的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

亲爱的所有人,





我的xml文件的rootnode为Base,我的属性Name可用在根节点本身。我需要找出位于我的根节点中的属性值。我该怎么办,请尽快帮助我。供您参考,我特此为您提供样品。



< Baseclass ID =5Name =Raj>

< nextclass>

< / Baseclass>

解决方案

试试这个

 XDocument xDoc = XDocument.Load(  xml的路径); 
var TagIds = xDoc.Descendants()
.Elements()
.Where(e = >
e.HasAttributes&&
e.Name.LocalName.Equals( 基类)&&
e.Attribute( ID)!= null
。选择(e = > e.Attribute( ID)。值);







希望这有助于


Dim tempPath As String = FolderBrowserDialog1.SelectedPath& \& ListBox1.Items(x).ToString



Dim xSid =来自xattr In xDoc.Descendants(Baseclass)_

选择att = xattr.Attribute(Name)。值

sSID = xSid.ElementAt(0).ToString''SID值

Dear All,


I have my xml file with rootnode as "Base" and I have my attribute "Name" available in the root node itself. I need to find out my attribute value located in my root node. How could I proceed, please help me ASAP. For your reference I hereby providing you a sample.

<Baseclass ID="5" Name="Raj">
<nextclass>
</Baseclass>

解决方案

Try this

XDocument xDoc = XDocument.Load("path to xml");
var TagIds = xDoc.Descendants()
                .Elements()
                .Where(e => 
                            e.HasAttributes && 
                            e.Name.LocalName.Equals("Baseclass") && 
                            e.Attribute("ID") != null)
                .Select(e => e.Attribute("ID").Value);




Hope this helps


Dim tempPath As String = FolderBrowserDialog1.SelectedPath & "\" & ListBox1.Items(x).ToString

Dim xSid = From xattr In xDoc.Descendants("Baseclass") _
Select att = xattr.Attribute("Name").Value
sSID = xSid.ElementAt(0).ToString ''SID value


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