c# open file dialog
OpenFileDialog dialog = new OpenFileDialog();
if (DialogResult.OK == dialog.ShowDialog())
{
string path = dialog.FileName;
}c# open file dialog
OpenFileDialog dialog = new OpenFileDialog();
if (DialogResult.OK == dialog.ShowDialog())
{
string path = dialog.FileName;
}open file dialog c#
var fileContent = string.Empty;
var filePath = string.Empty;
using (OpenFileDialog openFileDialog = new OpenFileDialog())
{
openFileDialog.InitialDirectory = "c:\";
openFileDialog.Filter = "txt files (*.txt)|*.txt|All files (*.*)|*.*";
openFileDialog.FilterIndex = 2;
openFileDialog.RestoreDirectory = true;
if (openFileDialog.ShowDialog() == DialogResult.OK)
{
//Get the path of specified file
filePath = openFileDialog.FileName;
//Read the contents of the file into a stream
var fileStream = openFileDialog.OpenFile();
using (StreamReader reader = new StreamReader(fileStream))
{
fileContent = reader.ReadToEnd();
}
}
}
MessageBox.Show(fileContent, "File Content at path: " + filePath, MessageBoxButtons.OK);open file dialog in vb.net
'How to open the file path in Visual Basic
OpenFileDialog1.Filter = "TEXT FILE | *.txt |PDF FILE |.pdf |ALL FILES |*.*"
If OpenFileDialog1.ShowDialog = DialogResult.OK Then
For x = 0 To OpenFileDialog1.FileNames.Count - 1
MsgBox(OpenFileDialog1.FileNames(x))
Next
End IfCopyright © 2021 Codeinu
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