WPF如何在ListView中为每个School生成对应DataGrid？

完全可以实现这个需求，你之前的XAML写法错误在于没有为ListView指定ItemTemplate——ListView的ItemsSource绑定的是Schools集合，每个列表项的上下文是单个School对象，而非整个集合，直接在ListView内部放DataGrid会导致绑定失效。

正确的XAML实现 #

<ListView ItemsSource="{Binding Schools}">
    <ListView.ItemTemplate>
        <DataTemplate>
            <!-- 用StackPanel垂直排列学校相关元素和教师表格 -->
            <StackPanel Margin="0 0 0 15">
                <!-- 可选：如果School类添加Name属性，可在此显示学校名称 -->
                <TextBlock Text="教师列表" FontSize="14" FontWeight="Bold" Margin="0 0 0 5"/>
                
                <DataGrid ItemsSource="{Binding Teachers}" AutoGenerateColumns="False">
                    <!-- 手动定义列，精准控制显示内容 -->
                    <DataGrid.Columns>
                        <DataGridTextColumn Header="姓名" Binding="{Binding Name}"/>
                        <DataGridTextColumn Header="ID" Binding="{Binding ID}"/>
                        <DataGridTextColumn Header="地址" Binding="{Binding Address}"/>
                        <DataGridTextColumn Header="排名" Binding="{Binding Ranking}"/>
                    </DataGrid.Columns>
                </DataGrid>
            </StackPanel>
        </DataTemplate>
    </ListView.ItemTemplate>
</ListView>

关键说明 #

• ListView.ItemTemplate：定义每个School项的渲染结构，模板内的DataContext是单个School实例，因此DataGrid的ItemsSource直接绑定Teachers即可
• AutoGenerateColumns="False"：关闭自动列生成，手动定义列能避免不必要的属性列，同时自定义表头和显示顺序
• 若需展示学校名称等额外信息，只需在School类中添加对应属性，再在StackPanel内新增TextBlock绑定该属性即可

测试用ViewModel初始化示例 #

为快速验证效果，可在MainViewModel构造函数中添加测试数据：

public class MainViewModel
{
    public ObservableCollection<School> Schools { get; set; }

    public MainViewModel()
    {
        Schools = new ObservableCollection<School>
        {
            new School
            {
                Teachers = new ObservableCollection<Teacher>
                {
                    new Teacher { Name = "张三", ID = "T001", Address = "北京", Ranking = "A" },
                    new Teacher { Name = "李四", ID = "T002", Address = "上海", Ranking = "B" }
                }
            },
            new School
            {
                Teachers = new ObservableCollection<Teacher>
                {
                    new Teacher { Name = "王五", ID = "T003", Address = "广州", Ranking = "A" }
                }
            }
        };
    }
}

内容的提问来源于stack exchange，提问作者zynovij