如何在C#中弹出窗口 [英] How to popup window in C#

问题描述

大家好，

我现在在Visual Studio中开发一个窗口窗体.我的项目中有两种形式.当我单击在Form1中显示按钮时，它将弹出form2，但是我做不到.请帮助我任何解决方案.

Hi all,

I am now developing a window form in Visual Studio. there are two form in my project. When I click button show in Form1 then It will popup the form2, but I cannot do it. Please help me any solution. Thank you in advance.

推荐答案

在Form1中按钮的按钮单击事件上，创建Form2的实例，请执行以下代码.

On button click event of button in Form1, make an instance of your Form2, do the below codes.

Form2 form2 = new Form2();
form2.Show();

如果这解决了您的问题，请标记为答案并投票5

问候，
爱德华

Please mark as answer and vote 5 if this solved your problem

Regards,
Eduard

在窗体1的按钮单击事件中输入以下代码

in form 1 button click event enter the following code

{
form2 frm = new form2(); //it will initialize the object frm for form 2 in form 1
frm.Show();// this will opening form 2

//if you want to hide form 1 then
this.Hide();
}

尝试一下.

在button_Click事件中，

try this.

inside the button_Click event,

form2.show();

希望对您有帮助

别忘了投票

谢谢

hope it helps

don''t forget to vote

thank

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