C#最简单写入读出XML配置文件

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总结下做项目时的技术点

首先看一下XML文件

<?xml version="1.0" encoding="utf-8"?><Soft>  <SN DataSource="192.168.0.115" DataName="shipdata" UserId="root" Password="abc" timer="2000"></SN></Soft>

Set.cs类

using System;using System.Collections.Generic;using System.Configuration;using System.Linq;using System.Text;using System.Xml;using System.Xml.Linq;namespace BatchQueryShip{public class Set{public static string FileUrl = "";/// <summary> /// 返回XMl文件指定元素的指定属性值 /// </summary> /// <param name="xmlElement">指定元素</param> /// <param name="xmlAttribute">指定属性</param> /// <returns></returns> public static string getXmlValue(string xmlElement, string xmlAttribute){ XDocument xmlDoc = XDocument.Load(FileUrl); var results = from c in xmlDoc.Descendants(xmlElement)select c; string s = ""; foreach (var result in results) { s = result.Attribute(xmlAttribute).Value.ToString(); } return s;} /// <summary> /// 设置XMl文件指定元素的指定属性的值 /// </summary> /// <param name="xmlElement">指定元素</param> /// <param name="xmlAttribute">指定属性</param> /// <param name="xmlValue">指定值</param> public static void setXmlValue( string xmlElement, string xmlAttribute, string xmlValue){ XDocument xmlDoc = XDocument.Load(FileUrl); xmlDoc.Element("Soft").Element(xmlElement).Attribute(xmlAttribute).SetValue(xmlValue); xmlDoc.Save(FileUrl); }}}

然后是先指定下XML的目录位置

Set.FileUrl = Application.StartupPath + "\\set.xml";

开始读XML数据了;

textBox1.Text = Set.getXmlValue("SN", "DataSource"); textBox1.Text = Set.getXmlValue("SN", "DataSource取xml数据写法    Set.setXmlValue("SN", "DataSource", textBox1.Text);

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