如何编写和读取xml文件的数据，以便在c＃中保存windows窗体的数据

本文介绍了如何编写和读取xml文件的数据，以便在c＃中保存windows窗体的数据的处理方法，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 我有一个Windows窗体包含datagridview（有学生信息）和菜单（有项目（保存，打开） i想要那个 $ b填写所有学生信息后$ b - >如果用户点击保存按钮 保存为窗口将打开 - >然后用户输入文件的名称，文件将保存（文件的扩展名可以是任何东西..例如.sgc） --->然后所有的数据用户输入的datagridview将保存在该文件上 与想要打开该文件相同.. 请帮忙.. !!这是代码.. private void saveAsToolStripMenuItem_Click（ object sender，EventArgs e） { string folderPath = string .Empty; SaveFileDialo g savefiledialog1 = new SaveFileDialog（）; savefiledialog1.Filter = Xml文件（* .xml）| * .xml |所有文件（*。 。*）| *。*; savefiledialog1.RestoreDirectory = true ; if （savefiledialog1.ShowDialog（）== DialogResult.OK） { folderPath = savefiledialog1.FileName; if （！System.IO.File.Exists（folderPath）） { XmlDeclaration declaration = doc.CreateXmlDeclaration（ 1.0， UTF-8， yes）; XmlComment comment = doc.CreateComment（ 这是一个XML生成的文件）; XmlElement root = doc.CreateElement（ calcs）; XmlElement calc = doc.CreateElement（ calc）; XmlAttribute Name = doc.CreateAttribute（ Name）; // 为每个节点添加值 Name.Value = textBox5。文本; // 构建文档 doc .AppendChild（声明）; doc.AppendChild（评论）; doc.AppendChild（root）; root.AppendChild（calc）; calc.Attributes.Append（Name）; doc.Save（savefiledialog1.FileName）; } // 显示确认消息 MessageBox.Show（ 详细信息已添加到XML文件中。）; // 重置新输入的文本字段 // textBox5.Text = String.Empty; } } private void openSignalChainToolStripMenuItem_Click（ object sender，EventArgs e） { XmlDataDocument xmldoc = new XmlDataDocument（）; OpenFileDialog openfiledialog1 = new OpenFileDialog（）; // openfiledialog1.InitialDirectory =c：\\; openfiledialog1.FileName = Document; openfiledialog1.DefaultExt = 。xml; openfiledialog1.Filter = xml file（.xml）| * .xml; openfiledialog1.FilterIndex = 2 ; Nullable< bool> result = Convert.ToBoolean（openfiledialog1.ShowDialog（））; openfiledialog1.RestoreDirectory = true ; if （result == true ） { string filename = openfiledialog1.FileName; 尝试 { XmlNodeList xmlnode; int i = 0 ; string str = null ; FileStream fs = new FileStream（filename，FileMode.Open，FileAccess.Read）; xmldoc.Load（fs）; xmlnode = xmldoc.GetElementsByTagName（ Product）; for （i = 0 ; i < ; = xmlnode.Count - 1 ; i ++） { xmlnode [i] .ChildNodes.Item（ 0 ）InnerText.Trim（）; str = xmlnode [i] .ChildNodes.Item（ 0 ）。InnerText.Trim（）+ | + xmlnode [i] .ChildNodes.Item（ 1 ）。InnerText.Trim（ ）+ | + xmlnode [i] .ChildNodes.Item（ 2 ）InnerText.Trim（）; MessageBox.Show（str）; textBox5.Text = str; } } catch （Exception ex） { MessageBox.Show（ 找不到文件!!!!）; } } } 解决方案 如果您使用DataSet作为您网格的数据，只需使用WriteXml保存数据，使用ReadXml将数据加载回网格。 http://msdn.microsoft.com/en-us/library/System.Data.DataSet ％28V = vs.110％29.aspx i have a windows form contains datagridview (having student information) and a menu(having items(save,open)i want thatafter filling all student information--> if user click save button save as window will open-->then user entered the name of file and file will save (extension of file could be anything ..for example .sgc)---> then all the data of datagridview that entered by user will save on that file same as a want to open that file..please help..!!!Here is code..private void saveAsToolStripMenuItem_Click(object sender, EventArgs e) { string folderPath = string.Empty; SaveFileDialog savefiledialog1 = new SaveFileDialog(); savefiledialog1.Filter = "Xml files (*.xml)|*.xml|All files (*.*)|*.*"; savefiledialog1.RestoreDirectory = true; if (savefiledialog1.ShowDialog() == DialogResult.OK) { folderPath = savefiledialog1.FileName; if (!System.IO.File.Exists(folderPath)) { XmlDeclaration declaration = doc.CreateXmlDeclaration("1.0", "UTF-8", "yes"); XmlComment comment = doc.CreateComment("This is an XML Generated File"); XmlElement root = doc.CreateElement("calcs"); XmlElement calc = doc.CreateElement("calc"); XmlAttribute Name = doc.CreateAttribute("Name"); //Add the values for each nodes Name.Value = textBox5.Text; //Construct the document doc.AppendChild(declaration); doc.AppendChild(comment); doc.AppendChild(root); root.AppendChild(calc); calc.Attributes.Append(Name); doc.Save(savefiledialog1.FileName); } //Show confirmation message MessageBox.Show("Details have been added to the XML File."); //Reset text fields for new input //textBox5.Text = String.Empty; } } private void openSignalChainToolStripMenuItem_Click(object sender, EventArgs e) { XmlDataDocument xmldoc = new XmlDataDocument(); OpenFileDialog openfiledialog1 = new OpenFileDialog(); // openfiledialog1.InitialDirectory = "c:\\"; openfiledialog1.FileName = "Document"; openfiledialog1.DefaultExt = ".xml"; openfiledialog1.Filter = "xml file (.xml)|*.xml"; openfiledialog1.FilterIndex = 2; Nullable<bool> result = Convert.ToBoolean(openfiledialog1.ShowDialog()); openfiledialog1.RestoreDirectory = true; if (result==true) { string filename = openfiledialog1.FileName; try { XmlNodeList xmlnode; int i = 0; string str = null; FileStream fs = new FileStream(filename, FileMode.Open, FileAccess.Read); xmldoc.Load(fs); xmlnode = xmldoc.GetElementsByTagName("Product"); for (i = 0; i <= xmlnode.Count - 1; i++) { xmlnode[i].ChildNodes.Item(0).InnerText.Trim(); str = xmlnode[i].ChildNodes.Item(0).InnerText.Trim() + " | " + xmlnode[i].ChildNodes.Item(1).InnerText.Trim() + " | " + xmlnode[i].ChildNodes.Item(2).InnerText.Trim(); MessageBox.Show(str); textBox5.Text = str; } } catch (Exception ex) { MessageBox.Show("File can not found!!!!"); } } } 解决方案 If you use DataSet as datasorce of yours grid just use WriteXml to save data and ReadXml to load data back to grid.http://msdn.microsoft.com/en-us/library/System.Data.DataSet%28v=vs.110%29.aspx 这篇关于如何编写和读取xml文件的数据，以便在c＃中保存windows窗体的数据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！