问题描述
表格 1 有一个按钮,可以让您进入表格 2.(按下表格 1 关闭表格 2 显示)
form 1 has a button which gets you to form 2.(When pressed form 1 closes form 2 shows)
form 2 有一个秒表,一旦你按下 form1 中的按钮就会自动启动
form 2 has a stopwatch which automatically starts once you ve pressed the button in form1
表单 2 也有一个按钮,可以让您返回表单 1.(按下表单 2 时会关闭表单 1 显示)
form 2 also has a button which gets you back to form 1.(when pressed form 2 closes form 1 shows)
问题是,当第二次在表单 1 和表单 2 之间切换时,会创建一个新的表单 2 窗口,这意味着秒表会重新启动.
The problem is that when switching btween form 1 to form 2 for the SECOND time a new form 2 window is created and that means that the stopwatch restarts.
我希望能够在表单之间切换,并且秒表从第一次登录开始就不停地工作,并且仅在我终止应用程序时停止
I want to be able to switch between the forms and the stopwatch will work non - stop since the very first log - in and stops only when i terminate the app
我对如何解决这个问题有想法,但我是 C# 新手,所以我不知道如何实现.尽量具体说明你的答案,如果可能的话,给我一些你的想法的编码示例.
I have ideas on how to fix that but im completely new in C# so i dont know how to make that happen.Try to be specific with your answers and give me some coded examples of your thoughts if that s possible.
推荐答案
问题是当你 Close() form2 时,里面的东西被破坏了.因此,您要么需要一个单独的 Thread 用于 StopWatch,或者您可以尝试将表单设置为在未显示时不可见.
The problem is that when you Close() form2, that the stuff in it gets destroyed. So you either need a seperate Thread for your StopWatch or you could try to set the form to invisible while it is not shown.
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