Question

我有一个PowerShell数组，如果我导出为XML，它将如下所示：

<?xml version="1.0"?>
<Objects>
   <Object>
      <Property Name="Server">Server1</Property>
      <Property Name="Issue">hard drive 0 failed</Property>
   </Object>
</Objects>

我想让这个数组在GUI中显示。在谷歌搜索之后，我发现可以使用WPF来达到这个目标，但我从来没有使用过它，所以我很难解决这个问题：

如何正确地将数组绑定到ViewList以获取自动创建的列并命名为属性名称（“Server”，“Issue”等）
我不需要编辑数据，但希望每行都有一个按钮，将当前行的数据作为新问题发送到Jira。我已经有一个代码如何使用PS和REST API创建Jira问题，但还不知道如何在ListView单元格中放置一个按钮

经过另一天的谷歌搜索后，我发现DataGrid更适合我的情况，因为根据传递的ItemsSource对象自动生成列数。

现在我遇到了一个新问题，这是我的代码：

$btn = New-Object System.Windows.Controls.Button
$dt = New-Object System.Windows.DataTemplate
$dt.VisualTree = New-Object System.Windows.FrameworkElementFactory($btn)
$dgtc = New-Object System.Windows.Controls.DataGridTemplateColumn
$dgtc.CellTemplate = $dt
$faults = Get-UcsFault -Severity critical | select @{n=’Server’; e={$_.Dn}},
          @{n=’Issue’;e={$_.Issue}}, 
          @{n=’Create Ticket’;e={[System.Windows.Controls.DataGridTemplateColumn]$dgtc}}
$result = New-Object System.Windows.Window -prop @{
   Content = New-Object System.Windows.Controls.DataGrid -prop @{ItemsSource=$faults};}
$result.ShowDialog()

但是我没有在实际按钮中看到最后一栏中的类名（即System.Windows.Controls.Button）

Answer 1

听起来你想根据XML对象创建一个类。

public class yourObject //Whatever you want to call it.
{
  public string Server {get;set;}
  public string Issue {get;set;}
  //ETC
}

完成类设置后，Id会浏览XML文件并将数据放入列表中。

        string filepath = Directory.GetParent(Environment.CurrentDirectory).Parent.FullName + @"\OBJECTS.xml"; 
       //Path to XML file

        List<yourObject> list = new List<yourObject>();
        var maps = from c in XElement.Load(filepath).Elements("Object")
                   select c;

        foreach (var item in maps)
        {
            var newObject= new yourObject();

            //Names of your nodes
            newObject.Server= item.Element("Server").Value;
            newObject.Issue= item.Element("Issue").Value;

            list.Add(newObject);
        }
        //Probably a good idea to make it an observable collection.
        objectList= new ObservableCollection<youtObject>();
        foreach (var item in list)
        {
            objectList.Add(item);
        }

一旦它在列表中，并且您不想绑定它，您只需将Datagrid的Item源或listview设置为该列表

Answer 2

这是一种使用Linq to XML在Code Behind of window中读取数据的方法：

public partial class MainWindow : Window
{
    private ObservableCollection<ServerIssue> serverIssues;

    public MainWindow()
    {

        // Linq to XML for elegant Selections
        XElement doc = XElement.Load("serverissues.xml");
        var xmlServerIssues = from objectElt in doc.Descendants("Object")
                           let propertiesElts = objectElt.Elements("Property")
                           let serverElt = propertiesElts.FirstOrDefault(elt => elt.Attribute("Name").Value == "Server")
                           let issueElt = propertiesElts.FirstOrDefault(elt => elt.Attribute("Name").Value == "Issue")
                           select new ServerIssue {
                               Server = ((serverElt.FirstNode) as XText).Value,
                               Issue = ((issueElt.FirstNode) as XText).Value
                           };
        // store enumerated data in ObservableCollection<>
        // List<> would be fine but is less agile when adding/removing elements
        serverIssues = new ObservableCollection<ServerIssue>(xmlServerIssues);
        InitializeComponent();
        // Provide data to graphical Grid
        datagridIssues.ItemsSource = serverIssues;
    }
    private void SendToJIRA(object sender, RoutedEventArgs e)
    {
        // ...
    }
}

DataGrid布局代码，绑定属性：

<Window x:Class="XmlToDataGrid.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">
    <Grid>
        <DataGrid x:Name="datagridIssues" CanUserAddRows="False" CanUserDeleteRows="False" AutoGenerateColumns="False">
            <DataGrid.Columns>
                <DataGridTextColumn Binding="{Binding Server}" IsReadOnly="true"/>
                <DataGridTextColumn Binding="{Binding Issue}" IsReadOnly="true"/>
                <DataGridTemplateColumn >
                    <DataGridTemplateColumn.CellTemplate>
                        <DataTemplate>
                            <Button Content="Send to JIRA" Click="SendToJIRA" Margin="10 4" Padding="5 3"/>
                        </DataTemplate>
                    </DataGridTemplateColumn.CellTemplate>
                </DataGridTemplateColumn>
            </DataGrid.Columns>
        </DataGrid>
    </Grid>
</Window>

修改

这是PowerShell中的翻译。即使在C＃中，它也不那么明显。我真的很鼓励你用C＃编写这样的代码。您将拥有一个编译器，一个严肃的调试器。

[reflection.assembly]::loadwithpartialname("PresentationFramework") | Out-Null # $faults = Get-UcsFault -Severity critical | select @{n=’Server’; e={$_.Dn}}, # @{n=’Issue’;e={$_.Issue}}, # @{n=’Create Ticket’;e={$bt}} ###################################################################################### $serverIssue1 = New-Object psobject -Property @{ Server="server1" Issue="issue1" } $serverIssue2 = New-Object psobject -Property @{ Server="server2" Issue="issue2" } $serverIssues = New-Object System.Collections.Generic.List``1[Object] $serverIssues.Add( $serverIssue1 ) $serverIssues.Add( $serverIssue2 ) $datagrid = New-Object Windows.Controls.DataGrid -prop @{ AutoGenerateColumns=$FALSE; ItemsSource = $serverIssues; CanUserAddRows=$FALSE CanUserDeleteRows=$FALSE } #$datagrid.ItemsSource=$faults # create the binding for a column $bindingIssue = New-Object System.Windows.Data.Binding -prop @{ Path='Issue'} # create the column $columnIssue = New-Object System.Windows.Controls.DataGridTextColumn -prop @{ IsReadOnly=$TRUE;Binding=$bindingIssue} $datagrid.Columns.Add($columnIssue) # create the binding for a column $bindingServer = New-Object System.Windows.Data.Binding -prop @{ Path='Server'} # create the column $columnServer = New-Object System.Windows.Controls.DataGridTextColumn -prop @{ IsReadOnly=$TRUE;Binding=$bindingServer} $datagrid.Columns.Add($columnServer) # create 3rd column : it s a template, more complicated ! Real WPF hardcore stuff # not easy in C#, hard tests in PS ! $factory = New-Object System.Windows.FrameworkElementFactory([System.Windows.Controls.Button]) $factory.SetValue( [System.Windows.Controls.Button]::ContentProperty, "Send to Jira") $thickness = New-Object System.Windows.Thickness(10,5,10,5) $factory.SetValue( [System.Windows.Controls.Button]::PaddingProperty, $thickness ) $factory.SetValue( [System.Windows.Controls.Button]::MarginProperty, $thickness ) # let's plug a callback when the button is clicked $factory.AddHandler([System.Windows.Controls.Button]::ClickEvent, [System.Windows.RoutedEventHandler]{ PARAM ([Object]$sender, [Windows.RoutedEventArgs]$e) # replace with code that sends to Jira # the DataContext is for the current line of DataGrid. It 's inherited by every component of the line # so the button has a serverIssue DataContext $serverIssue = $sender.DataContext [System.Windows.MessageBox]::Show($serverIssue.Server); }) $dataTemplate = New-Object System.Windows.DataTemplate -prop @{ VisualTree = $factory } $columnButton = New-Object System.Windows.Controls.DataGridTemplateColumn -prop @{ IsReadOnly = $TRUE; CellTemplate = $dataTemplate } $datagrid.Columns.Add($columnButton) $result = New-Object System.Windows.Window -prop @{ Content = $datagrid } $result.ShowDialog()

此致

将数组/ XML绑定到WPF ListView / DataGrid

2 个答案: