Can datagridview read a hyperlink?

Question

The excel sheet's hyperlink I have will direct to a folder in the user's pc. What I want to do is import from excel the hyperlink to the DataGridView and from there, be able to open the file the hyperlink directs to. However, I keep seeing people setting the hyperlink website inside their c# code which is not what I'm trying to find since it's hard coding it. Is there a way for it to be achieved?

Answer 1

It's not something automatic and you need to code it manually. You need to use a DataGridViewLinkColumn, and handle its CellContentClick and then pass the value of the cell as url to System.Diagnostics.Process.Start method.

If you have a lot of locations in your application which you have such a requirement, you can make a function to reuse the logic or create custom DataGridViewLinkColumn to have this behavior inside.

Example

private void Form1_Load(object sender, EventArgs e)
{
    //Get data
    var data = new[] {
        new { Title="Stackoverflow", Location = @"https://www.Stackoverflow.com"},
        new { Title="Windows Folder", Location = @"C:\Windows"},
        new { Title="Windows Folder", Location = @"C:\Windows\system.ini"},
        new { Title="Network Share", Location = @"\\127.0.0.1"},
    };

    // Add columns
    dataGridView1.Columns.Add(new DataGridViewTextBoxColumn()
    {
        Name = "TitleColumn", DataPropertyName = "Title", HeaderText = "Title"
    });
    dataGridView1.Columns.Add(new DataGridViewLinkColumn()
    {
        Name = "LocationColumn", DataPropertyName = "Location", HeaderText = "Location"
    });

    //Handle click on link
    dataGridView1.CellContentClick += (obj, args) =>
    {
        if (args.RowIndex < 0 || args.ColumnIndex < 0)
            return;
        if (dataGridView1.Columns[args.ColumnIndex].Name != "LocationColumn")
            return;
        var value = $"{dataGridView1[args.ColumnIndex, args.RowIndex].Value}";
        if (Uri.TryCreate(value, UriKind.Absolute, out Uri uri))
            System.Diagnostics.Process.Start(value);
    };

    //Show data
    dataGridView1.DataSource = data.ToList();
}