在c#中从xml获取属性值

Question

我正在使用 C# 并且我想解析关系元素。
我想获得“MarketplaceId”（A1F83G8C2ARO7P）和“ASIN”（B076B1GP37）的价值

这是我的 XML。

   <Relationships>
     <VariationParent xmlns="http://mws.amazonservices.com/schema/Products/2011-10-01">
      <Identifiers>
        <MarketplaceASIN>
          <MarketplaceId>A1F83G8C2ARO7P</MarketplaceId>
          <ASIN>B076B1GP37</ASIN>
        </MarketplaceASIN>
      </Identifiers>
    </VariationParent>
  </Relationships>

到目前为止，这是我的代码。

if (relationshipList.IsSetAny())
{
   foreach(var relationship in relationshipList.Any)
   {
      string rxml = ProductsUtil.FormatXml((System.Xml.XmlElement)relationship);
      XDocument xDoc = XDocument.Parse(rxml);
      XNamespace ns = XNamespace.Get("http://mws.amazonservices.com/schema/Products/2011-10-01");

      IEnumerable<object> relationships = xDoc.Descendants();

      foreach (System.Xml.Linq.XElement xe in relationships)
      {
         string r_marketplaceId = (string)xe.Attribute("MarketplaceId");
         string r_ASIN = (string)xe.Attribute("ASIN");                                        
      }
    }
}

xe 上面是下面

 <VariationParent xmlns="http://mws.amazonservices.com/schema/Products/2011-10-01">
  <Identifiers>
    <MarketplaceASIN>
      <MarketplaceId>A1F83G8C2ARO7P</MarketplaceId>
      <ASIN>B076B1GP37</ASIN>
    </MarketplaceASIN>
  </Identifiers>
</VariationParent>

r_marketplaceId 和 r_ASIN 仍然为空值.. 任何建议和建议将不胜感激。

Answer 1

我怀疑问题是 - 您没有在这里处理属性。 考虑示例：

<VariationParent xmlns="http://mws.amazonservices.com/schema/Products/2011-10-01">
  <Identifiers>
    <MarketplaceASIN MarketplaceId="A1F83G8C2ARO7P" ASIN="B076B1GP37" /> <!-- these are attributes -->
  </Identifiers>
</VariationParent>
<!------------>
 <VariationParent xmlns="http://mws.amazonservices.com/schema/Products/2011-10-01">
  <Identifiers>
    <MarketplaceASIN>
      <MarketplaceId>A1F83G8C2ARO7P</MarketplaceId> <!-- these are values -->
      <ASIN>B076B1GP37</ASIN> <!-- these are values -->
    </MarketplaceASIN>
  </Identifiers>
</VariationParent>

第一个片段依赖于attributes而你的则将数据表示为values 。 因此，您可能需要像这样更改代码：

string r_marketplaceId = (string)xe.XPathSelectElement("//MarketplaceId").Value;
string r_ASIN = (string)xe.XPathSelectElement("//ASIN").Value;

希望这能解决你的问题

Answer 2

第一点，您正在声明xn并且不使用它。

您可以通过LINQ查询获取值，如下面的代码：

XNamespace xn = "http://mws.amazonservices.com/schema/Products/2011-10-01";

IEnumerable<string> elements = XDocument.Parse(rxml)
    .Descendants(xn + "MarketplaceASIN")
    .Descendants()
    .Where(e=>!e.HasElements)
    .Select(element => element.Value);

Console.WriteLine(string.Join(", ", elements));

结果

A1F83G8C2ARO7P、B076B1GP37

我希望这会帮助你。

Answer 3

将 xml linq 与字典一起使用。 见下面的代码

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Xml;
using System.Xml.Linq;

namespace ConsoleApplication159
{
    class Program
    {
        const string FILENAME = @"c:\temp\test.xml";
        static void Main(string[] args)
        {
            XDocument doc = XDocument.Load(FILENAME);
            XNamespace ns = doc.Root.GetDefaultNamespace();
            Dictionary<string, string> dict1 = doc.Descendants(ns + "MarketplaceASIN")
                .GroupBy(x => (string)x.Element(ns + "MarketplaceId"), y => (string)y.Element(ns + "ASIN"))
                .ToDictionary(x => x.Key, y => y.FirstOrDefault());
            //where MarketplaceId may be repeated
            Dictionary<string, List<string>> dict2 = doc.Descendants(ns + "MarketplaceASIN")
                 .GroupBy(x => (string)x.Element(ns + "MarketplaceId"), y => (string)y.Element(ns + "ASIN"))
                .ToDictionary(x => x.Key, y => y.ToList());
        }
    }
}