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 [英] How Do I write and read data of xml file for save the data of windows form in c#

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< /跨度>）; 
 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 that

after 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

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