C＃从XML属性获取值 [英] C# get values from xml attributes

问题描述

如何使用C＃来获得属性行动和文件名的价值观正确的方式

How to get attribute "action" and "filename" values in a right way using C#?

XML：

<?xml version="1.0" encoding="utf-8" ?>
 <Config version="1.0.1.1" >
   <Items>
    <Item action="Create" filename="newtest.xml"/>
    <Item action="Update" filename="oldtest.xml"/>   
  </Items>
 </Config>

C＃：我不能得到属性值，以及如何获取值在fo​​reach循环？ ？如何解决这个

C#: i cannot get attribute values as well as how to get values in foreach loops? How to solve this?

        var doc = new XmlDocument();
        doc.Load(@newFile);
        var element = ((XmlElement)doc.GetElementsByTagName("Config/Items/Item")[0]); //null
        var xmlActions = element.GetAttribute("action"); //cannot get values
        var xmlFileNames= element.GetAttribute("filename"); //cannot get values

         foreach (action in xmlActions)
         {
           //not working
         }

         foreach (file in xmlFileNames)
         {
           //not working
         }

您的代码例如很多手段对我。谢谢！

Your code example means alot to me. Thanks!

推荐答案

您可以使用 LINQ到XML 。下面的查询返回强类型项目的集合，动作和文件名属性：

You can use LINQ to XML. Following query returns strongly typed collection of items with Action and FileName properties:

var xdoc = XDocument.Load(@newFile);

var items = from i in xdoc.Descendants("Item")
            select new {
               Action = (string)i.Attribute("action"),
               FileName = (string)i.Attribute("fileName")
            };

foreach (var item in items)
{
   // use item.Action or item.FileName
}

这篇关于C＃从XML属性获取值的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！