I have a console application in which we are creating xlsx files using OPENXML, we are able to create xlsx file & save it into specific folder in application.
But Now we want to show that file as a Save/Open dialog pop up. Then we can able to specify a particular path to save/ to open the existing files.
I am new to this OpenXml, Can anyone please help me on this to proceed further? How can I acheive this? Do we have any built-in DLL for this?
Thanks.
se the Save file dialog. It will prompts the user to select a location for saving a file. After that you can use saveFileDialog.FileName.ToString() property to get the full path.
See the sample code below:
//Save a file in a particular format as specified in the saveAsType parameter
private void OpenSaveFileDialog(int saveAsType)
{
SaveFileDialog saveFileDialog = new SaveFileDialog();
saveFileDialog.InitialDirectory = Convert.ToString(Environment.SpecialFolder.MyDocuments);
saveFileDialog.Filter = "CSV|*.csv|Excel|*.xlsx";
saveFileDialog.FilterIndex = saveAsType;
saveFileDialog.Title = "Save Data";
saveFileDialog.FileName = "My File";
saveFileDialog.ShowDialog();
if (saveFileDialog.FileName != "")
{
//File Path = m_fileName
m_fileName = saveFileDialog.FileName.ToString();
//FilterIndex property is one-based.
switch (saveFileDialog.FilterIndex)
{
case 1:
m_fileType = 1;
break;
case 2:
m_fileType = 2;
break;
}
}
}
Ref:http://msdn.microsoft.com/en-us//library/system.windows.forms.savefiledialog.aspx
