C# 如何在ListView上仅显示项目的属性
我已经在ListView上添加了一个类“Employee”的对象列表,我希望它只显示属性“Name”,但当选择该项时,我希望它捕获要添加到另一个列表中的对象“Employee” 这是我的密码。它在ListView上显示对象Employee。我希望它显示属性名称C# 如何在ListView上仅显示项目的属性,c#,wpf,listview,C#,Wpf,Listview,我已经在ListView上添加了一个类“Employee”的对象列表,我希望它只显示属性“Name”,但当选择该项时,我希望它捕获要添加到另一个列表中的对象“Employee” 这是我的密码。它在ListView上显示对象Employee。我希望它显示属性名称 private void PopulateEmployeeList() { List<Employee> staff = Presenter.GetEmployee(); for
private void PopulateEmployeeList()
{
List<Employee> staff = Presenter.GetEmployee();
foreach (Employee person in staff)
lstEmployee.Items.Add(person);
}
private void PopulateEmployeeList()
{
List staff=Presenter.GetEmployee();
foreach(员工中的员工)
lstmemployee.Items.Add(个人);
}
XAML
您可以为ListView定义ItemContainerStyle:
<ListView Name="lstFuncionarios">
<ListView.ItemContainerStyle>
<Style TargetType="{x:Type ListViewItem}" >
<Setter Property="Template">
<Setter.Value>
<ControlTemplate TargetType="{x:Type ListViewItem}">
<TextBlock Text="{Binding Name}" />
</ControlTemplate>
</Setter.Value>
</Setter>
</Style>
</ListView.ItemContainerStyle>
</ListView>
或
希望有帮助您可以在xaml中尝试:
<ListView x:Name="lv" HorizontalAlignment="Left" Height="149" Margin="359,111,0,0" VerticalAlignment="Top" Width="138">
<ListView.View>
<GridView>
<GridViewColumn DisplayMemberBinding="{Binding Name}"/>
</GridView>
</ListView.View>
</ListView>
查看此链接。请同时提供您的XAML。XAML仅包含ListView的名称。我想您正在查找DisplayMemberBinding
lstFuncionarios.DisplayMemberPath = "Name";
<ListView Name="lstFuncionarios" DisplayMemberPath="Name"/>
<ListView x:Name="lv" HorizontalAlignment="Left" Height="149" Margin="359,111,0,0" VerticalAlignment="Top" Width="138">
<ListView.View>
<GridView>
<GridViewColumn DisplayMemberBinding="{Binding Name}"/>
</GridView>
</ListView.View>
</ListView>