Question

我有以下xml文件来生成我们网站的菜单。

 <xs:element name="Menu">
    <xs:complexType>
        <xs:sequence>
            <xs:element name="MenuItem" type="MenuItemType" maxOccurs="unbounded"></xs:element>
        </xs:sequence>
        <xs:attribute name="Title" type="xs:string"></xs:attribute>
        <xs:attribute name="Type" type="xs:string"></xs:attribute>
    </xs:complexType>

</xs:element>
<xs:complexType name="MenuItemType">
    <xs:choice minOccurs="0" maxOccurs="unbounded">
        <xs:element name="MenuItem" type="MenuItemType" />
    </xs:choice>
    <xs:attribute name="Text" type="xs:string"></xs:attribute>
    <xs:attribute name="Url" type="xs:string"></xs:attribute>
</xs:complexType>

现在我正在使用xmlserializer将这些xml文件转换为Menu对象，并使用它们来生成菜单。我想使用LINQ to xml将这些xml文件转换为同一个对象。任何帮助将不胜感激。上面的xml文件的生成类是

 public partial class Menu {

    /// <remarks/>
    [System.Xml.Serialization.XmlElementAttribute("MenuItem")]
    public MenuItemType[] MenuItem;

    /// <remarks/>
    [System.Xml.Serialization.XmlAttributeAttribute()]
    public string Title;
    /// <remarks/>
    [System.Xml.Serialization.XmlAttributeAttribute()]
    public string Type;
}
public partial class MenuItemType {

    /// <remarks/>
    [System.Xml.Serialization.XmlElementAttribute("MenuItem")]
    public MenuItemType[] Items;

    /// <remarks/>
    [System.Xml.Serialization.XmlAttributeAttribute()]
    public string Text;

    /// <remarks/>
    [System.Xml.Serialization.XmlAttributeAttribute()]
    public string Url;
}

Answer 1

我还没有测试过。但是，希望这有效。

var o = (from e in XDocument.Load("").Elements("MenuItem")
         select new Menu
         {
             MenuItem = GenerateMenuItemType(e).ToArray(),
             Title = (string)e.Attribute("Title"),
             Type = (string)e.Attribute("Type")
         });

private IEnumerable<MenuItemType> GenerateMenuItemType(XElement element)
{
    return (from e in element.Elements("MenuItem")
            select new MenuItemType
            {
                Items = GenerateMenuItemType(e).ToArray(),
                Text = (string)e.Attribute("Title"),
                Url = (string)e.Attribute("Url")
            });
}

如何使用LINQ将xml文件转换为对象

1 个答案: