Question

我有以下XML文档

 <Documents Databasename="Type1">
<Document>
  <Documentnumber>xxxxxx</Documentnumber>
  <Documentname>xxxxxx</Documentname>
  <DocumentExtension>xxxxxx</DocumentExtension>
  <ImportAction>xxxxxx</ImportAction>
  <Authorname>xxxxxx</Authorname>
  <Practicegroup>xxxxxx</Practicegroup>
  <Createddate>xxxxxx</Createddate>
  <Createdby>xxxxxx</Createdby>
  <Modifieddate>xxxxxx</Modifieddate>
  <Modifiedby>xxxxxx</Modifiedby>
  <Lastaccessddate>xxxxxx</Lastaccessddate>
  <Comments />
  <Sector>Unknown</Sector>
  <Source>Unknown</Source>
  <Quality>Unknown</Quality>
  <KnowHowContentType>Unknown</KnowHowContentType>
  <ActualAutors />
  <SubmittedBy />
  <NumberofAccess>0</NumberofAccess>
  <AreasofLaw />
</Document>
<Document>
  <Documentnumber>xxxxxx</Documentnumber>
  <Documentname>xxxxxx</Documentname>
  <DocumentExtension>xxxxxx</DocumentExtension>
  <ImportAction>xxxxxx</ImportAction>
  <Authorname>xxxxxx</Authorname>
  <Practicegroup>xxxxxx</Practicegroup>
  <Createddate>xxxxxx</Createddate>
  <Createdby>xxxxxx</Createdby>
  <Modifieddate>xxxxxx</Modifieddate>
  <Modifiedby>xxxxxx</Modifiedby>
  <Lastaccessddate>xxxxxx</Lastaccessddate>
  <Comments />
  <Sector>Unknown</Sector>
  <Source>Unknown</Source>
  <Quality>Unknown</Quality>
  <KnowHowContentType>Unknown</KnowHowContentType>
  <ActualAutors />
  <SubmittedBy />
  <NumberofAccess>0</NumberofAccess>
  <AreasofLaw />
</Document>

XXXXXX XXXXXX XXXXXX XXXXXX XXXXXX XXXXXX XXXXXX XXXXXX XXXXXX XXXXXX XXXXXX 未知未知未知未知 0 XXXXXX XXXXXX XXXXXX XXXXXX XXXXXX XXXXXX XXXXXX XXXXXX XXXXXX XXXXXX XXXXXX 未知未知未知未知 0

我想为给定属性Databasename选择DocumentNumber，DocumentExtension。如果我解释了自己，请告诉我。

Answer 1

我没有太多使用Linq到XML，所以这可能不是最好的方法，但似乎有效：

string xml = @"
<root>
    <Documents Databasename='Type1'>
    <Document>
      <Documentnumber>1234</Documentnumber>";
      // etc..

XDocument xmlDoc = XDocument.Parse(xml);

var output = from item in xmlDoc.Descendants()
             where item.Name == "Document"
             && item.Parent.Attribute("Databasename").Value == "Type1"
             select new { Number = (string)item.Element("Documentnumber").Value,
                          Extension = item.Element("DocumentExtension").Value
             };

foreach (var item in output) {
    Console.WriteLine(item);
}

请注意，我在XML中添加了一个额外的“root”元素以使其工作（假设有多个“Documents”元素）。

LinqXML通过属性过滤器 - 初学者

1 个答案: