我正在使用wpf。我想绑定一个文本框,其中包含在xaml.cs类中初始化的简单字符串类型值。 TextBox
没有显示任何内容。这是我的XAML代码:
<TextBox Grid.Column="1" Width="387" HorizontalAlignment="Left" Grid.ColumnSpan="2" Text="{Binding Path=Name2}"/>
C#代码是这样的:
public partial class EntitiesView : UserControl
{
private string _name2;
public string Name2
{
get { return _name2; }
set { _name2 = "abcdef"; }
}
public EntitiesView()
{
InitializeComponent();
}
}
答案 0 :(得分:7)
您永远不会设置您的财产的价值。在实际执行set操作之前,简单地定义set { _name2 = "abcdef"; }
实际上并不会设置属性的值。
您可以将代码更改为以下内容:
public partial class EntitiesView : UserControl
{
private string _name2;
public string Name2
{
get { return _name2; }
set { _name2 = value; }
}
public EntitiesView()
{
Name2 = "abcdef";
DataContext = this;
InitializeComponent();
}
}
此外,正如人们所提到的,如果您打算稍后修改属性的值并希望UI反映它,则需要实现INotifyPropertyChanged
接口:
public partial class EntitiesView : UserControl, INotifyPropertyChanged
{
private string _name2;
public string Name2
{
get { return _name2; }
set
{
_name2 = value;
RaisePropertyChanged("Name2");
}
}
public EntitiesView()
{
Name2 = "abcdef";
DataContext = this;
InitializeComponent();
}
public event PropertyChangedEventHandler PropertyChanged;
protected void RaisePropertyChanged(string propertyName)
{
var handler = PropertyChanged;
if (handler != null)
{
handler(this, new PropertyChangedEventArgs(propertyName));
}
}
}
答案 1 :(得分:2)
只需在EntitiesView
构造函数
DataContext = this;
答案 2 :(得分:0)
为什么不添加视图模型并将属性保留在那里?
查看模型类
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.ComponentModel;
namespace WpfApplication1
{
public class TestViewModel : INotifyPropertyChanged
{
public string _name2;
public string Name2
{
get { return "_name2"; }
set
{
_name2 = value;
OnPropertyChanged(new PropertyChangedEventArgs("Name2"));
}
}
public event PropertyChangedEventHandler PropertyChanged;
public void OnPropertyChanged(PropertyChangedEventArgs e)
{
if (PropertyChanged != null)
{
PropertyChanged(this, e);
}
}
}
}
EntitiesView用户控件
public partial class EntitiesView : UserControl
{
public EntitiesView()
{
InitializeComponent();
this.DataContext = new TestViewModel();
}
}