c# - 如何使用linq从xml中获取具有相同名称的不同值

Question

XML - 代码:

<Store>
   <Products>
      <Product id="PROD01">
         <Title>Product 1</Title>
         <Description><![CDATA[Product <b>1</b> description]]></Description>
         <Image>prod01.gif</Image>
         <Specs>
            <Spec>Good computer</Spec>
            <Spec>Good display</Spec>
            <Spec>Latest version</Spec>
         </Specs>
         <Availability>same day</Availability>
      </Product>
      <Product id="PROD02">
         <Title>Product 2</Title>
         <Description><![CDATA[Product <b>2</b> description]]></Description>
         <Image>prod01.gif</Image>
         <Specs>
            <Spec>Good computer</Spec>
            <Spec>Soon available</Spec>
         </Specs>
         <Availability>next day</Availability>
      </Product>
   </Products>
</Store>

C# - 代码:

public List<DetailList> GetDetails()
{
    DetailList d = new DetailList();
    List<DetailList> DetailLists = 
        (from product in xdocList.Descendants("Product")
        join detail in xdocDetail.Descendants("Product")
        on (string)product.Attribute("id") equals (string)detail.Attribute("id") 
        into outerProducts
        from outerProduct in outerProducts

        select new DetailList
        {
            Detail1 = (string)product.Attribute("id"),
            Detail2 = (string)product.Element("Title"),
            Detail3 = (string)product.Element("Description"),
            Detail4 = (string)product.Element("Image"),
            Detail5 = (string)outerProduct.Elements("Specs")
            Detail6 = (string)outerProduct.Element("Availability"),
            Detail7 = (string)product.Element("Price"),
        }).ToList();
    return DetailLists;
}

输出:好的电脑好的显示器最新版本

但想要的输出是:

好电脑
显示效果不错
最新版本

对于输出，我使用了 asp:repeater。我尝试添加 < b r/> 等标签，但找不到我的错误，如何将 Spec 获取到三个不同的字符串，而不仅仅是一个字符串。如何实现？

Answer 1

我不确定你为什么要用自己加入节点，但根据你的 XML，这不是必需的。您可以像这样简单地投影元素:-

public static List<DetailList> GetDetails(XDocument xdocList)
{
    DetailList d = new DetailList();
    List<DetailList> DetailLists = (from product in xdocList.Descendants("Product")
                                    select new DetailList
                                    {
                                        Detail1 = ((string)product.Attribute("id")),
                                        Detail2 = ((string)product.Element("Title")),
                                        Detail3 = ((string)product.Element("Description")),
                                        Detail4 = ((string)product.Element("Image")),
                                        Detail5 = product.Element("Specs")
                                                   .Elements("Spec")
                                                   .Select(x => (string)x).ToList(),
                                        Detail6 = ((string)product.Element("Availability")),
                                        Detail7 = ((string)product.Element("Price")),
                                     }).ToList();
            return DetailLists;
}

因为您需要单独的所有Specs，所以您应该有一个字符串集合，而不仅仅是字符串。所以属性 Detail5 的数据类型应该是:-

List<string> or string[]