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Qivorch Velix
Asked:2020-05-19 13:27:26 +0000 UTC2020-05-19 13:27:26 +0000 UTC

定时器按钮显示

  • 772

我遇到了一个问题：我需要一个带有动作的按钮在启动程序后一段时间后出现，但我做不到（我刚开始学习所有这些编程的东西，所以我请你找出简单的可能的）。在这里，我请求帮助，下面是按钮代码。

private void Button1_Click_1(object sender, EventArgs e)
        {
            foreach (var process in Process.GetProcessesByName("Thing"))
            {
                process.Kill();
            }

        }
        private void Button1_Paint(object sender, PaintEventArgs e)
        {
            button1.Hide();
            var stateTimer = new System.Timers.Timer(10);
            stateTimer.Elapsed += OnTimeEvent;
        }
        private void OnTimeEvent(Object source, ElapsedEventArgs e)
        {
            button1.Show();
        }
c#

1 个回答

  1. Best Answer
      private void Form1_Load(object sender, EventArgs e)
        {
            var stateTimer = new System.Timers.Timer(); // таймер
            stateTimer.Elapsed += ShowBtn; // подписаваемся на событие 
            stateTimer.Interval = 10000; // интервал покажет кнопку через 10секунд
            stateTimer.Start(); // запуск таймера
            button1.Visible = false;

        }

    void ShowBtn(object sender, EventArgs e)
    {
        if (InvokeRequired)
        {
            Invoke(new MethodInvoker(delegate
            {
                button1.Visible = true; // показать кнопку

            }));
        }

    }
    • 0

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