我正在为Windows 8.1开发一个应用程序。我能够将XML文件解析为ListView。它是本地XML文件,但是当我将XML路径(XMLFile1.xml)作为URL(例如http://192.168.x.x/test.xml)时,它会给我一个错误。 这是我的代码。
XAML文件:
<ListView x:Name="listView1" HorizontalAlignment="Left" Height="373" Margin="1229,264,0,0" VerticalAlignment="Top" Width="127">
<ListView.ItemTemplate>
<DataTemplate>
<StackPanel Margin="10" >
<TextBlock Text="{Binding FirstName}"/>
<TextBlock Text="{Binding LastName}"/>
<TextBlock Text="{Binding Age}"/>
</StackPanel>
</DataTemplate>
</ListView.ItemTemplate>
</ListView>
XAML.cs文件:
public class Person
{
string firstname;
string lastname;
int age;
public string FirstName
{
get { return firstname; }
set { firstname = value; }
}
public string LastName
{
get { return lastname; }
set { lastname = value; }
}
public int Age
{
get { return age; }
set { age = value; }
}
}
private void Page_Loaded(object sender, RoutedEventArgs e)
{
string peopleXMLPath = Path.Combine(Package.Current.InstalledLocation.Path, "XMLFile1.xml");
XDocument loadedData = XDocument.Load(peopleXMLPath);
var data = from query in loadedData.Descendants("person")
select new Person
{
FirstName = (string)query.Element("firstname"),
LastName = (string)query.Element("lastname"),
Age = (int)query.Element("age")
};
listView1.ItemsSource = data;
}
答案 0 :(得分:1)
您需要列出尝试通过外部源加载数据时使用的代码。如果您尚未更改此设置,请尝试以下选项。
尝试更改
XDocument loadedData = XDocument.Load(peopleXMLPath);
到
XDocument loadedData = XDocument.Load("http://192.168.x.x/test.xml");