WPF 窗口 ShowDialog() 导致无法设置可见性或调用 Show
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原文地址: http://stackoverflow.com/questions/19070991/
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WPF Window ShowDialog() causing Cannot set Visibility or call Show
提问by Jim
I am creating a wpf form which is going to be used for adding/editing data from datagrid. However when I check for ShowDialog() == trueI am getting the above exception.
我正在创建一个 wpf 表单，它将用于从 datagrid 添加/编辑数据。但是，当我检查ShowDialog() == true时，出现上述异常。
The code is taken from a book (Windows Presentation Foundation 4.5 Cookbook).
该代码取自一本书（Windows Presentation Foundation 4.5 Cookbook）。
UserWindow usrw = new UserWindow();
usrw.ShowDialog();
if (usrw.ShowDialog() == true)
{
//do some stuff here;
}
And on the WPF window:
在 WPF 窗口中：
private void btn_Save_Click(object sender, RoutedEventArgs e)
{
DialogResult = true;
Close();
}
How I can handle this?
我该如何处理？
===============================
================================
The solution to the problem was simply to remove usrw.ShowDialog(); and it start working as expected
解决问题的方法很简单，就是去掉 usrw.ShowDialog(); 它开始按预期工作
UserWindow usrw = new UserWindow();
//usrw.ShowDialog();
if (usrw.ShowDialog() == true)
{
//do some stuff here;
}
回答by Florian
You are trying to open your window 2 times with every call to ShowDialog()
每次调用时，您都试图打开窗口 2 次 ShowDialog()
try
尝试
UserWindow usrw = new UserWindow();
bool result =(bool)usrw.ShowDialog();
if (result)
{
//do some stuff here;
}
or
或者
UserWindow usrw = new UserWindow();
usrw.ShowDialog();
if ((bool)usrw.DialogResult)
{
//do some stuff here;
}
keep in mind that DialogResultis Nullable. If there is a chance that you are closing the window without setting the DialogResult, check for null.
请记住，它DialogResult是可空的。如果您有可能在没有设置 DialogResult 的情况下关闭窗口，请检查null.
