Visual Studio C ++ 2005中表单的链接 [英] linking of forms in visual studio c++2005

问题描述

i hav 2通过位于form1中的按钮从form1调用Form2，从而形成Form1和Form2.i.使用下面的代码form2将打开，但是form1仍然可见.


Form2().ShowDialog();
Form2().Activate();
Form1().Close();

我只希望可见form2.当form2处于活动状态时如何隐藏form1?pls help ..

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@Mohammad Elsheimy如果我包含Form2().Show()而不是Form2().ShowDialog()form1打开，但是form2打开jus占一秒.我希望当我运行我的应用程序时，form2应该在我单击到一个菜单时打开

解决方案

您的应用程序将无法进入第二行:
Form2().Activate();
直到完成第一个ShowDialog()为止.

ShowDialog函数将阻止您的代码，直到窗体关闭

您应该将其更改为
Form2().Show();

i hav 2 forms Form1 n Form2.i called form2 from the form1 thru a button which is in form1. Using the following code form2 opens but form1 is still visible.


Form2().ShowDialog();
Form2().Activate();
Form1().Close();

i want only form2 to be visible .how can i hide form1 when form2 is active?pls help..

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@Mohammad Elsheimy if i include Form2().Show() instead of Form2().ShowDialog()form1 opens but form2 opens jus for a fraction of second.i want that when i run my application form2 should open when i click to a button in form1 and form1 should be invisible.

解决方案

Your application won''t get to the 2nd line:
Form2().Activate();
until it completes the first ShowDialog().

ShowDialog function blocks your code until the form closes

you should change it to
Form2().Show();

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