Open file from shared drive path on ASP.Net LinkButton Click using C# and VB.Net

Hi Tevin,

Use Process.Start to open the file from shared folder.

Refer  below sample.

HTML

<asp:GridView ID="gvDetails" runat="server" AutoGenerateColumns="false">
    <Columns>
        <asp:BoundField DataField="Text" HeaderText="FileName" />
        <asp:TemplateField>
            <ItemTemplate>
                <asp:LinkButton ID="lnkOpen" runat="server" Text='<%#DataBinder.Eval(Container.DataItem, "Text")%>'
                    CommandArgument='<%#DataBinder.Eval(Container.DataItem, "Value")%>' OnClick="OnOpen"></asp:LinkButton>
            </ItemTemplate>
        </asp:TemplateField>
    </Columns>
</asp:GridView>

Namespaces

C#

using System.Diagnostics;
using System.IO;

VB.Net

Imports System.Diagnostics
Imports System.IO

Code

C#

protected void Page_Load(object sender, EventArgs e)
{
    if (!this.IsPostBack)
    {
        string[] folderPaths = Directory.GetFiles(@"Shared Folder Path");
        List<ListItem> files = new List<ListItem>();
        foreach (string filePath in folderPaths)
        {
            files.Add(new ListItem(Path.GetFileName(filePath), filePath));
        }
        gvDetails.DataSource = files;
        gvDetails.DataBind();
    }
}

protected void OnOpen(object sender, EventArgs e)
{
    string filePath = (sender as LinkButton).CommandArgument;
    Process.Start(filePath);
}

VB.Net

Protected Sub Page_Load(ByVal sender As Object, ByVal e As EventArgs) Handles Me.Load
    If Not Me.IsPostBack Then
        Dim folderPaths As String() = Directory.GetFiles("Shared Folder Path")
        Dim files As List(Of ListItem) = New List(Of ListItem)()

        For Each filePath As String In folderPaths
            files.Add(New ListItem(Path.GetFileName(filePath), filePath))
        Next

        gvDetails.DataSource = files
        gvDetails.DataBind()
    End If
End Sub

Protected Sub OnOpen(ByVal sender As Object, ByVal e As EventArgs)
    Dim filePath As String = (TryCast(sender, LinkButton)).CommandArgument
    Process.Start(filePath)
End Sub

Screenshot