现在我有这段代码从MDIParent窗口打开名为MDIParent1的图像
private void OpenFile(object sender, EventArgs e)
{
OpenFileDialog openFileDialog = new OpenFileDialog();
openFileDialog.InitialDirectory = Environment.GetFolderPath(Environment.SpecialFolder.Personal);
openFileDialog.Filter = "Image Files (*.jpg)|*.jpg|All Files (*.*)|*.*";
if (openFileDialog.ShowDialog(this) == DialogResult.OK)
{
string FileName = openFileDialog.FileName;
Process.Start(@FileName);
}
}
这打开了一个新的窗口,我的图像很好,但我希望它作为MDIParent1的子窗口打开。任何帮助都将非常感激。谢谢你
Process.Start(@FileName);
与在资源管理器中双击文件相同。它会打开一个新窗口。如果您设置了默认程序来打开图像,那么它将打开程序。
如果你想用c#来做,创建一个带有PictureBox的表单。然后不调用Process.Start(@FileName);
,而是像这样调用表单:
Form1 form = new Form1();
form.MdiParent = this;
form.PictureBox1.ImageLocation = FileName;
form.Open();