Creating a Context Menu on a DataGridView Mouse Click

This article describes how to create a context menu when you mouse click on a DataGridView control in Windows Forms.

One of the projects had a requirement to create a context menu on mouse click over a DataGridView having employee details. The menu items may vary from column to column of the gridview.

//Define different context menus for different columns
private ContextMenu contextMenuForColumn1 = new ContextMenu();
private ContextMenu contextMenuForColumn2 = new ContextMenu();

Add the following line of code in the form load event:

private void Form_Load(object sender, EventArgs e)
{
    // Load all default values of controls
    populateDataGridView();

    // Add context mneu items
    contextMenuForColumn1.MenuItems.Add("Make Active", new     EventHandler(MakeActive));
    contextMenuForColumn2.MenuItems.Add("Delete", new     EventHandler(Delete));
    contextMenuForColumn2.MenuItems.Add("Register", new     EventHandler(Register));
}

Add the following code to mouseup event of the gridview:

private void dataGridView_MouseUp(object sender, MouseEventArgs e)
{
    // Load context menu on right mouse click
    DataGridView.HitTestInfo hitTestInfo;
    if (e.Button == MouseButtons.Right)
    {
        hitTestInfo = dataGridView.HitTest(e.X, e.Y);
        // If column is first column
        if (hitTestInfo.Type == DataGridViewHitTestType.Cell && hitTestInfo.ColumnIndex == 0)
            contextMenuForColumn1.Show(dataGridView, new Point(e.X, e.Y));
        // If column is second column
        if (hitTestInfo.Type == DataGridViewHitTestType.Cell && hitTestInfo.ColumnIndex == 1)
            contextMenuForColumn2.Show(dataGridView, new Point(e.X, e.Y));
    }
}