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“c# open file dialog” Code Answer’s

c# open file dialog

csharp by Ugly Unicorn on Jul 16 2020 Donate

OpenFileDialog dialog = new OpenFileDialog();
if (DialogResult.OK == dialog.ShowDialog()) 
{
    string path = dialog.FileName;
}

open file dialog c#

csharp by Calm Crocodile on Sep 27 2020 Donate

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);

Source: docs.microsoft.com