使用列将项添加到listview行

时间:2012-10-13 11:16:16

标签: c# wpf listview

所以我试图在我的列表视图中添加一行信息,但是当我这样做时,显示它很奇怪。像这样:

enter image description here

我正在为每个循环使用a:

foreach (Client c in clients)
{
   ListViewItem i = new ListViewItem();
   i.Content = new String[] { c.info.cid.ToString(), c.info.pc.ToString(),c.info.ip.ToString(), c.info.status.ToString() };
   list.Items.Add(i);
}

我的客户端类使用结构来存储信息

public struct Info
{
  public int cid;
  public string pc;
  public string ip;
  public string status;
}

我也在为它添加值:

info = new Info();
info.ip = "192.168.1.100";
info.pc = "Duncan";
info.status = "idle";
info.cid = 1;

为什么它显得怪异?有人可以帮忙吗?

我的ListView XAML:

<ListView Height="247" HorizontalAlignment="Left" Margin="4,6,0,0" Name="list" VerticalAlignment="Top" Width="319" Background="#FF454545" ItemsSource="{Binding}" SelectionMode="Multiple" Grid.Column="0">
   <ListView.View>
       <GridView AllowsColumnReorder="False">
       <GridViewColumn Header="ID" Width="30" />
       <GridViewColumn Header="Computer" Width="100" />
       <GridViewColumn Header="IP" Width="100" />
       <GridViewColumn Header="Status" Width="100" />
       </GridView>
    </ListView.View>
    </ListView>

1 个答案:

答案 0 :(得分:2)

此代码中存在一些错误的内容。如果要使用绑定在ListView中推送数据,则必须具有要绑定的属性的有效ViewModel。您必须在GridViewColumns上定义绑定。

此外,WPF不知道如何绑定字段,因此您需要为要显示的每个数据使用.NET属性。这是一个非常原始的例子,它不是一个现实的场景,但应该可以帮助你开始:

<强> Window.xaml

<Window x:Class="WpfApplication1.MainWindow"
    xmlns="http://schemas.microsoft.com/winfx/2006/xaml/presentation"
    xmlns:x="http://schemas.microsoft.com/winfx/2006/xaml"
    Title="MainWindow" Height="350" Width="525">
<ListView Height="247" HorizontalAlignment="Left" Margin="4,6,0,0" Name="list" VerticalAlignment="Top" Width="319" Background="#FF454545" ItemsSource="{Binding Clients}" SelectionMode="Multiple" Grid.Column="0">
    <ListView.View>
        <GridView AllowsColumnReorder="False">
            <GridViewColumn Header="ID" Width="30" DisplayMemberBinding="{Binding Id}" />
            <GridViewColumn Header="Computer" Width="100" DisplayMemberBinding="{Binding Computer}" />
            <GridViewColumn Header="IP" Width="100" DisplayMemberBinding="{Binding Ip}" />
            <GridViewColumn Header="Status" Width="100" DisplayMemberBinding="{Binding Status}" />
        </GridView>
    </ListView.View>
</ListView>
</Window>

<强> MainWindow.xaml.cs

/// <summary>
/// Logique d'interaction pour MainWindow.xaml
/// </summary>
public partial class MainWindow : Window
{
    public MainWindow()
    {
        InitializeComponent();
        DataContext = this;           
    }

    public IEnumerable<DummyClient> Clients
    {
        get
        {
            for (int i = 0; i < 10; i++)
            {
                var info = new Info();
                info.ip = "192.168.1.100";
                info.pc = "Duncan";
                info.status = "idle";
                info.cid = 1;

                yield return new DummyClient(info);
            }
        }
    }
}

public class DummyClient
{
    public DummyClient(Info info)
    {
        Info = info;
    }

    public string Ip { get { return Info.ip; } }
    public string Computer { get { return Info.pc; } }
    public string Status { get { return Info.status; } }
    public int Id { get { return Info.cid; } }

    public Info Info
    {
        get;
        private set;
    }
}

public struct Info
{
    public int cid;
    public string pc;
    public string ip;
    public string status;
}

再一次,它不是应该做的方式,但这是一个开始。例如,如果你想要两种方式绑定工作,DummyClient应该实现INotifyPropertyChanged。