Make GUI Button (client) With ServerEvent

FiveM Resource Development & Modding Discussion
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Hello FiveM forums, If I create this topic is that I find no solution to my problem. Explanation!

In my LUA menu I want to display the information present in one of my MYSQL tables;

Except that many attempts not functional I look on yours help!

I recover with success the information of my table but I do not know how to convert them afins these appear as a button in my menu LUA

(Sorry for my english google translation)

Here is my code now, If you could help me with the complete

server/server.lua

RegisterNetEvent('journaliste:helicoptaire_menu_vehicle')
AddEventHandler('journaliste:helicoptaire_menu_vehicle', function()
    local source = source
    MySQL.Async.fetchAll("SELECT * FROM entreprise_vehicle WHERE entreprise_id = @entreprise_id AND vehicle_type = @vehicle_type", { ['@entreprise_id'] = entreprise_journaliste_id, ['@vehicle_type'] = 'helico' }, function(result)
        if result ~= nil then
            for k, v in pairs(result) do
                print('open menu  helicoptaire_menu_vehicle - ' .. result[k].vehicle_spawn_name .. ' ID :' .. result[k].id)
            end
        end
    end)
end)

client/client.lua (Menu LUA)

function MenuWeazelNewsHelicoptaire()
    options.menu_subtitle = 'Garage Helicoptaire : Weazel News'
    ClearMenu()
    TriggerServerEvent('journaliste:helicoptaire_menu_vehicle')

    Menu.addButton('~r~Fermer ', "CloseMenu", nil)
end

Thank you for your future help

Discord : iTexZoz - Dylan M.#4280