我在c#中更新WPF表单中的TextBox
文本时遇到问题。我已经以编程方式创建了新表单并添加了一个Label
和一个TextBox
,并且我有一个变量temp,它表示我之前创建的某个缓冲区中的某个字符串。但是,当我尝试将文本设置为标签或文本框时,没有任何反应。但是,我可以用新的形式更改窗口标题。
我的代码是:
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading;
using System.Threading.Tasks;
using System.Windows;
using System.Windows.Controls;
using System.Windows.Data;
using System.Windows.Documents;
using System.Windows.Input;
using System.Windows.Media;
using System.Windows.Media.Imaging;
using System.Windows.Navigation;
using System.Windows.Shapes;
namespace BufferProba
{
/// <summary>
/// Interaction logic for MainWindow.xaml
/// </summary>
public partial class MainWindow : Window
{
BufferStream bfStream = new BufferStream();
private static Action EmptyDelegate = delegate() { };
public MainWindow()
{
InitializeComponent();
}
private void Window_Loaded(object sender, RoutedEventArgs e)
{
Thread t = new Thread(SetText);
t.SetApartmentState(ApartmentState.STA);
t.IsBackground = true;
t.Start();
}
private void Button_Click_1(object sender, RoutedEventArgs e)
{
bfStream.put(tBox.Text.Trim());
tBox.Text = "";
}
public void SetText()
{
Thread.Sleep(5000);
Window myWindow = new Window();
StackPanel stackPanel = new StackPanel { Orientation = Orientation.Vertical };
TextBox tboxForm = new TextBox();
Label szzr = new Label { Content = "" };
stackPanel.Children.Add(szzr);
stackPanel.Children.Add(tboxForm);
myWindow.Content = stackPanel;
List<String> listaStringova = new List<String>();
while (true)
{
Thread.Sleep(5000);
String temp = bfStream.get();
listaStringova.Add(temp);
if (temp != "0")
{
//Console.WriteLine(temp);
myWindow.Title = temp;
szzr.Content = temp;
szzr.Background = new SolidColorBrush(Colors.Orange);
szzr.UpdateLayout();
tboxForm.Text = temp;
myWindow.Show();
}
else {
MessageBox.Show("Jebiga");
}
}
}
}
}
答案 0 :(得分:0)
您需要确保它是访问TextBox的UI线程。使用Dispatcher.CheckAccess方法,如果返回false,则BeginInvoke UI线程。您应该能够在网上找到大量有关此信息的示例。
答案 1 :(得分:0)
您应该在主线程上运行它。 您可以使用Dispatcher执行相同的操作 例如:
System.Windows.Application.Current.Dispatcher.Invoke(new Action(SetText()))