WinForm 读写XML文件

2008-07-25 16:13:18.0     推荐:0    收藏:0    评论:0     来源:中国IT实验室

建立一个WinForm应用程序 添加MenuStrip控件,填写两个功能“读取” 和“导出数据”。

用了两个DataSet控件和对话框“打开(OpenFilesDialog控件)"和"保存(SaveFilesDialog控件)"

读取 private void 读取ToolStripMenuItem_Click(object sender, EventArgs e)
{
if (opFileDlg .ShowDialog() == DialogResult.OK)
{
if(opFileDlg .OpenFile()!=null)
{
twoXML .ReadXml (@opFileDlg .FileName );
foreach (DataRow twoRow in twoXML .Tables ["user"].Rows)
{
DataRow newRow = dsXML.Tables["user"].NewRow();
newRow ["序号"] = twoRow ["序号"];
newRow["标题"] = twoRow["标题"];
newRow["网址"] = twoRow["网址"];
newRow["用户名"] = twoRow["用户名"];
newRow["密码"] = twoRow["密码"];
newRow["时间"] = twoRow["时间"];
newRow["备注"] = twoRow["备注"];
dsXML .Tables ["user"].Rows .Add(newRow);
}
int n = dsXML .Tables ["user"].Rows .Count ;
for(int i=0;i {
dsXML .Tables ["user"].Rows [i]["序号"]=i+1;
}
dsXML.WriteXml(@"user.xml");
this.Visible = true;
MessageBox.Show("数据导入成功!", "成功");
}
}
else
{
this.Visible = true;
}
}

导出 private void 导出ToolStripMenuItem_Click(object sender, EventArgs e)
{
if (svFileDlg.ShowDialog() == DialogResult.OK)
{
dsXML.WriteXml(@svFileDlg.FileName);
this.Visible = true;
MessageBox.Show("数据导出成功!", "成功");
}
else
{
this.Visible = true;
}
}

您可以针对本文进行:[评论]  [收藏]  [推荐]  
  • 共有0条评论  点击查看更多评论
  • 网友评论仅供网友表达个人看法,并不表明e800同意其观点或证实其描述
我想发表评论:
用户名密码
  • 匿名发表
    验证码: