关闭在任务栏属性显示的表单设置为False。
it1352
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问题说明
我正在使用c＃开发一个Windows应用程序。我打开了一个表单并单击表单1的按钮，我打开表单2，即打开表单1和表单2。任务栏属性中的Form2显示设置为false。我想在表单1的按钮点击事件中关闭或隐藏表单2.我尝试使用隐藏和关闭方法，但仍然显示表单2.
任何帮助!!!
在此先感谢!!!
I am developing a windows application using c#. I have one form open and on button click of form 1, I am opening form 2, i.e. both Forms form 1 and form 2 are opened. Form2 Show in task-bar property is set to false. I want to close or hide the form 2 on button click event of form 1. I tried to use Hide and Close Method, but the form 2 is still displayed.
Any Help !!!
Thanks In Advance !!!
bool Display; // declared at class level.
//btnshow_hide is on form 1.
private void btnshow_hide_Click(object sender, EventArgs e)
{
form2 tc = new form2();
if (Display == true)
{
Display = false;
tc.Show();
}
else if (Display == false)
{
Display = true;
tc.Hide();
//tc.Close();
}
}正确答案
#1
您必须在表单的实际实例上使用表单的隐藏或关闭方法，而不是在新表单上使用：
You have to use the Hide or Close method of the form on the actual instance of the form, not on a new one:
private Form2 form2;
private void OpenForm2()
{
if (form2 == null)
{
form2 = new Form2();
}
form2.Show();
}
private void CloseForm2()
{
if (form2 != null)
{
form2.Close();
form2 = null;
}
}
< b>它无法正常工作。我在表单1上有按钮，我想使用相同的按钮来显示和隐藏表单2.
并不是一个很大的改变：
"Its not working. I have button on form 1, I want to use the same button to show and hide the form 2."
Not exactly a big change:
private Form2 form2;
private void ToggleForm2()
{
if (form2 == null)
{
form2 = new Form2();
form2.Show();
}
else
{
form2.Close();
form2 = null;
}
}这篇好文章是转载于：学新通技术网
