Question

我是XPath的新手。我需要帮助解决在本XML结束时提取Book的标题和作者所需的XPath。我尝试了以下C＃代码但没有成功。我只需要列出本书的标题和作者。看起来xmlns名称空间会影响我的XPath。如果我手动删除xmlns，我的代码可以正常工作。因此，要么修改XPath以考虑此命名空间，要么找出从XML中删除该属性的方法。请指教。

这是C＃代码：

XmlNodeList nodes = XML.DocumentElement.SelectNodes("//Title");
foreach (XmlNode node in nodes)
{
Console.WriteLn(node.Name + " = " + node.InnerText); }
}

这是XML：

<?xml version="1.0"?>
<ItemSearchResponse xmlns="http://webservices.amazon.com/AWSECommerceService/2011-08-01">
  <OperationRequest>
    <RequestId>xxxxx</RequestId>
    <Arguments>
      <Argument Name="Condition" Value="All"></Argument>
      <Argument Name="ResponseGroup" Value="Small,Images"></Argument>
      <Argument Name="SearchIndex" Value="Books"></Argument>
    </Arguments>
    <RequestProcessingTime>0.0735170000000000</RequestProcessingTime>
  </OperationRequest>
  <Items>
    <Request>
      <IsValid>True</IsValid>
      <ItemSearchRequest>
        <Condition>All</Condition>
        <Keywords>Perl</Keywords>
        <ResponseGroup>Small</ResponseGroup>
        <ResponseGroup>Images</ResponseGroup>
        <SearchIndex>Books</SearchIndex>
      </ItemSearchRequest>
    </Request>
    <TotalResults>3761</TotalResults>
    <TotalPages>377</TotalPages>
    <MoreSearchResultsUrl>http://www.amazon.com/gp/redirect.html?camp=2025&amp;creative=386001&amp;location=http%3A%2F%2Fwww.amazon.com%2Fgp%2Fsearch%3Fkeywords%3DPerl%26url%3Dsearch-alias%253Dstripbooks&amp;linkCode=xm2&amp;tag=geo01d-20&amp;SubscriptionId=AKIAJJBQEKP2X72RQ6XA</MoreSearchResultsUrl>
    <Item>
      <ASIN>1449303587</ASIN>
      <DetailPageURL>http://www.amazon.com/Learning-Perl-Randal-L-Schwartz/dp/1449303587%3FSubscriptionId%3DAKIAJJBQEKP2X72RQ6XA%26tag%3Dgeo01d-20%26linkCode%3Dxm2%26camp%3D2025%26creative%3D165953%26creativeASIN%3D1449303587</DetailPageURL>
      <ItemLinks>
        <ItemLink>
          <Description>Technical Details</Description>
          <URL>http://www.amazon.com/Learning-Perl-Randal-L-Schwartz/dp/tech-data/1449303587%3FSubscriptionId%3DAKIAJJBQEKP2X72RQ6XA%26tag%3Dgeo01d-20%26linkCode%3Dxm2%26camp%3D2025%26creative%3D386001%26creativeASIN%3D1449303587</URL>
        </ItemLink>
        <ItemLink>
          <Description>All Offers</Description>
          <URL>http://www.amazon.com/gp/offer-listing/1449303587%3FSubscriptionId%3DAKIAJJBQEKP2X72RQ6XA%26tag%3Dgeo01d-20%26linkCode%3Dxm2%26camp%3D2025%26creative%3D386001%26creativeASIN%3D1449303587</URL>
        </ItemLink>
      </ItemLinks>
      <SmallImage>
        <URL>http://ecx.images-amazon.com/images/I/51kTNE8aIXL._SL75_.jpg</URL>
        <Height Units="pixels">75</Height>
        <Width Units="pixels">58</Width>
      </SmallImage>
      <MediumImage>
        <URL>http://ecx.images-amazon.com/images/I/51kTNE8aIXL._SL160_.jpg</URL>
        <Height Units="pixels">160</Height>
        <Width Units="pixels">123</Width>
      </MediumImage>
      <LargeImage>
        <URL>http://ecx.images-amazon.com/images/I/51kTNE8aIXL.jpg</URL>
        <Height Units="pixels">500</Height>
        <Width Units="pixels">385</Width>
      </LargeImage>
      <ImageSets>
        <ImageSet Category="primary">
          <SwatchImage>
            <URL>http://ecx.images-amazon.com/images/I/51kTNE8aIXL._SL30_.jpg</URL>
            <Height Units="pixels">30</Height>
            <Width Units="pixels">23</Width>
          </SwatchImage>
          <SmallImage>
            <URL>http://ecx.images-amazon.com/images/I/51kTNE8aIXL._SL75_.jpg</URL>
            <Height Units="pixels">75</Height>
            <Width Units="pixels">58</Width>
          </SmallImage>
        </ImageSet>
      </ImageSets>
      <ItemAttributes>
        <Author>Randal L. Schwartz</Author>
        <Author>brian d foy</Author>
        <Author>Tom Phoenix</Author>
        <Manufacturer>O'Reilly Media</Manufacturer>
        <ProductGroup>Book</ProductGroup>
        <Title>Learning Perl</Title>
      </ItemAttributes>
    </Item>
  </Items>
</ItemSearchResponse>

Answer 1

您可以使用Linq-to-XML执行此操作：

XDocument xmlDoc = XDocument.Parse(xmlString);
var q = from el in xmlDoc.Descendants()
                  .Where(x => x.Name.LocalName == "Title" || x.Name.LocalName == "Author") 
                   select el;

foreach (var xElement in q)
{
    Debug.WriteLine(xElement.Name.LocalName + " : " + xElement.Value);
}

输出：

Author : Randal L. Schwartz
Author : brian d foy
Author : Tom Phoenix
Title : Learning Perl

Answer 2

XPath需要是＆＃34; // p：Title＆＃34;，你需要告诉XPath处理器命名空间前缀＆＃34; p＆＃34;代表命名空间＆＃34; http://webservices.amazon.com/AWSECommerceService/2011-08-01＆＃34;。建立命名空间绑定的方式取决于所选XPath处理器的API。对于C＃，您可以在此处找到解释（或者在许多其他以前的StackOverflow答案中）：

Xml-SelectNodes with default-namespace via XmlNamespaceManager not working as expected

Answer 3

我终于想出了我的问题的答案。我只是删除了过分复杂我的XPath的命名空间。下面的C＃代码有效。 PrintKeyValue只打印Key = Value。

            XML = new XmlDocument();

            using (XmlTextReader reader = new XmlTextReader("C:\\Path\\File.xml"))
            {
                reader.Namespaces = false;
                XML.Load(reader);
            }

            XmlNodeList items = XML.DocumentElement.SelectNodes("//Item");
            foreach (XmlNode item in items)
            {
                PrintKeyValue("ISBN10", item["ASIN"].InnerText);

                XmlNode attrib = item.SelectSingleNode(".//ItemAttributes");
                PrintKeyValue("Title", attrib["Title"].InnerText);

                XmlNodeList authors = attrib.SelectNodes(".//Author");
                StringBuilder sb = new StringBuilder();
                int count = 0;
                foreach (XmlNode author in authors)
                {
                    count++;
                    if (count > 1) { sb.Append(", "); }
                    sb.Append(author.InnerText);
                }
                PrintKeyValue("Authors", sb.ToString());
            }

带有命名空间的XML的XPath是什么？

3 个答案: