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我有一个看起来像这样的 xml 文件：

技术.xml

<?xml version="1.0" encoding="utf-8" ?>
<Tech>
<item ID="1" CATEGORY="example_1" NAME="example1" MIN_LEVEL1 ="1"    MAX_LEVEL1 ="10"/>

.

.

<item ID="102" CATEGORY="example_1222" NAME="example1333" MIN_LEVEL1 ="1"    MAX_LEVEL1 ="10"/>

</Tech>

我使用 xmldataprovider 将它绑定到数据网格的 DataGridTextColumn

<XmlDataProvider x:Key="tech" Source="tech.xml" XPath="Tech" />

像这样

<DataGrid ItemsSource="{Binding someotherclass}">

<DataGridTextColumn Binding  Source="{StaticResource tech}"   XPath="@NAME" />

.
.
.
other datagrid columns...

</DataGrid>

我遇到的问题是 DataGridTextColumn 输出中的所有行都显示值“example1”

而不是人们所期望的值 example1....example1333。

请注意，datagrid ItemsSource 绑定到其他一些数据文件，只有 DataGridTextColumn 绑定到 tech.xml。

4

1 回答 1

1

一种可能的方法 - 使用转换器。但是您需要有一些字段或属性，它将您的xml数据与someotherclass数据连接起来。我创建了通过索引连接两个集合的示例应用程序。希望它的帮助。

示例应用

这里的代码：

MainWindow.xaml：

<Window x:Class="XmlAndSeparateSource.MainWindow"
    xmlns="http://schemas.microsoft.com/winfx/2006/xaml/presentation"
    xmlns:x="http://schemas.microsoft.com/winfx/2006/xaml"
    xmlns:local="clr-namespace:XmlAndSeparateSource"
    Title="MainWindow" Height="350" Width="525">
<Window.Resources>        
    <XmlDataProvider x:Key="TechSource" Source="tech.xml"  XPath="Tech/item" />
    <local:RowNumberConverter x:Key="rowNumberConverter" />
</Window.Resources>
<Grid>
    <DataGrid ItemsSource="{Binding }"
              AutoGenerateColumns="False">
        <DataGrid.Columns>
            <DataGridTextColumn Header="xml">
                <DataGridTextColumn.Binding>
                    <MultiBinding Converter="{StaticResource rowNumberConverter}">
                        <Binding Source="{StaticResource TechSource}" />
                        <Binding RelativeSource="{RelativeSource FindAncestor, AncestorType={x:Type DataGrid}}" />
                        <Binding />
                    </MultiBinding>
                </DataGridTextColumn.Binding>
            </DataGridTextColumn>
            <DataGridTextColumn Header="string" Binding="{Binding}" />
        </DataGrid.Columns>
    </DataGrid>
</Grid>
</Window>

主窗口.cs

namespace XmlAndSeparateSource
{
/// <summary>
/// Interaction logic for MainWindow.xaml
/// </summary>
public partial class MainWindow : Window
{
    public MainWindow()
    {
        DataContext = new List<string>()
        {
            "str1",
            "str2",
            "str3",
            "str4",
        };
        InitializeComponent();

    }
}
}

RowNumberConverter.cs：

class RowNumberConverter : IMultiValueConverter
{
    public object Convert(object[] values, Type targetType, object parameter, System.Globalization.CultureInfo culture)
    {

        //get the grid and the item
        IEnumerable<XmlNode> xml = values[0] as IEnumerable<XmlNode>;
        DataGrid grid = values[1] as DataGrid;
        Object item = values[2];
        int index = grid.Items.IndexOf(item);

        return xml.ElementAt(index).Attributes.GetNamedItem("NAME").Value;
    }

    public object[] ConvertBack(object value, Type[] targetTypes, object parameter, System.Globalization.CultureInfo culture)
    {
        throw new NotImplementedException();
    }
}

技术.xml：

<Tech>
    <item ID="1" CATEGORY="example_1" NAME="example1" MIN_LEVEL1 ="1" MAX_LEVEL1 ="10"/>
    <item ID="2" CATEGORY="example_2" NAME="example2" MIN_LEVEL1 ="1" MAX_LEVEL1 ="10"/>
    <item ID="3" CATEGORY="example_3" NAME="example3" MIN_LEVEL1 ="1" MAX_LEVEL1 ="10"/>
    <item ID="102" CATEGORY="example_1222" NAME="example1333" MIN_LEVEL1 ="1" MAX_LEVEL1 ="10"/>
</Tech>
于 2012-04-25T08:54:49.773 回答