有关ShowDialog的问题

问题描述

大家好！我有一个关于ShowDialog方法的快速问题。我有第二个表单，我希望在单击按钮时显示。到目前为止我有这种方法：

Hi everyone! I have a quick question regarding the ShowDialog method. I have a second form that I want to be shown upon a button click. I have this method so far:

private void insertMakeButton_Click(object sender, EventArgs e)
        {
            newMakeForm.ActiveForm.ShowDialog();
        }

当我点击相应按钮时我得到的错误是：

已经可见的表单不能显示为模式对话框。在调用showDialog之前将表单的可见属性设置为false。



有人能说清楚我做错了吗？



谢谢大家！

The error I''m getting when I click on the corresponding button is:
Form that is already visible cannot be displayed as a modal dialog box. Set the form''s visible property to false before calling showDialog.

Can someone shed some light as to what I''m doing wrong?

Thanks everyone!

推荐答案

if( newMakeForm.Visible ) 
{ 
   newMakeForm.ActiveForm.Visible = false; 
}
newMakeForm.ActiveForm.ShowDialog();

这篇关于有关ShowDialog的问题的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！