通过使用 xpath 在 C# 中过滤现有文档来创建新的 XMLDocument

发布于 2024-09-26 22:40:52 字数 1678 浏览 5 评论 0原文

我遇到过这样的情况：我从外部公司收到 XML（文档）文件。我需要过滤文档以删除所有我不感兴趣的数据。该文件大约 500KB，但会经常被请求。

让我们说以下文件：

<dvdlist>
  <dvd>
    <title>title 1</title>
    <director>directory 2</director>
    <price>1</price>
    <location>
      <city>denver</city>
    </location>
  </dvd>
  <dvd>
    <title>title 2</title>
    <director>directory 2</director>
    <price>2</price>
    <location>
      <city>london</city>
    </location>
  </dvd>
  <dvd>
    <title>title 3</title>
    <director>directory 3</director>
    <price>3</price>
    <location>
      <city>london</city>
    </location>
  </dvd>
</dvdlist>

我需要的只是根据 city = london 过滤文档，以便最终得到这个新的 XML 文档

<dvdlist>
  <dvd>
    <title>title 2</title>
    <director>directory 2</director>
    <price>2</price>
    <location>
      <city>london</city>
    </location>
  </dvd>
  <dvd>
    <title>title 3</title>
    <director>directory 3</director>
    <price>3</price>
    <location>
      <city>london</city>
    </location>
  </dvd>
</dvdlist>

我已经尝试过以下

XmlDocument doc = new XmlDocument();
doc.Load(@"C:\Development\Website\dvds.xml");
XmlNode node = doc.SelectSingleNode("dvdlist/dvd/location/city[text()='london']");

任何帮助或链接，

谢谢

原文

I have a situation where I receive an XML (document) file from an external company. I need to filter the document to remove all data I am not interested in.
The file is about 500KB but will be requested very often.

let say the following file:

<dvdlist>
  <dvd>
    <title>title 1</title>
    <director>directory 2</director>
    <price>1</price>
    <location>
      <city>denver</city>
    </location>
  </dvd>
  <dvd>
    <title>title 2</title>
    <director>directory 2</director>
    <price>2</price>
    <location>
      <city>london</city>
    </location>
  </dvd>
  <dvd>
    <title>title 3</title>
    <director>directory 3</director>
    <price>3</price>
    <location>
      <city>london</city>
    </location>
  </dvd>
</dvdlist>

What I need is simply filter the document based on the city = london in order to end up with this new XML document

<dvdlist>
  <dvd>
    <title>title 2</title>
    <director>directory 2</director>
    <price>2</price>
    <location>
      <city>london</city>
    </location>
  </dvd>
  <dvd>
    <title>title 3</title>
    <director>directory 3</director>
    <price>3</price>
    <location>
      <city>london</city>
    </location>
  </dvd>
</dvdlist>

I have tried the following

XmlDocument doc = new XmlDocument();
doc.Load(@"C:\Development\Website\dvds.xml");
XmlNode node = doc.SelectSingleNode("dvdlist/dvd/location/city[text()='london']");

Any help or links will appreciate

Thanks

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李白 2024-10-03 22:40:52

XPath 是一种选择表达式语言——它从不修改它所操作的 XML 文档。

因此，为了获得所需的新 XML 文档，您需要使用 XML DOM（不推荐）或对 XML 文档应用 XSLT 转换。后者是推荐的方法，因为 XSLT 是一种专门为树转换而设计的语言。

在 .NET 中，可以使用 XslCompiledTransform 类 及其 Transform() 方法。在相关 MSDN 文档中了解有关这些内容的更多信息。

XSLT 转换本身非常简单：

<xsl:stylesheet version="1.0"
 xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
 <xsl:output omit-xml-declaration="yes" indent="yes"/>

 <xsl:template match="node()|@*">
  <xsl:copy>
   <xsl:apply-templates select="node()|@*"/>
  </xsl:copy>
 </xsl:template>

 <xsl:template match="dvd[not(location/city='london')]"/>
</xsl:stylesheet>

在这里，您可以找到完整的代码示例，如何以 XmlDocument 形式获取转换结果（或者如果需要，以 XDocument 形式）。

XPath is a selection expression language -- it never modifies the XML document(s) it operates on.

Therefore, in order to obtain the desired new XML document, you need to either use XML DOM (not recommended) or apply an XSLT transformation to the XML document. The latter is the recommended way to go, since XSLT is a language especially designed for tree transformations.

In .NET one can use the XslCompiledTransform class and its Transform() method. Read more about these in the relevant MSDN documentation.

The XSLT transformation itself is extremely simple:

<xsl:stylesheet version="1.0"
 xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
 <xsl:output omit-xml-declaration="yes" indent="yes"/>

 <xsl:template match="node()|@*">
  <xsl:copy>
   <xsl:apply-templates select="node()|@*"/>
  </xsl:copy>
 </xsl:template>

 <xsl:template match="dvd[not(location/city='london')]"/>
</xsl:stylesheet>

Here, you can find a complete code example how to obtain the result of the transformation as an XmlDocument (or if desired, as an XDocument).

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眼眸里的快感 2024-10-03 22:40:52

下面是使用 LINQ to XML 的示例。

//load the document
var document = XDocument.Load(@"C:\Development\Website\dvds.xml");
//get all dvd nodes
var dvds = document.Descendants().Where(node => node.Name == "dvd");
//get all dvd nodes that have a city node with a value of "london"
var londonDVDs = dvds.Where(dvd => dvd.Descendants().Any(child => child.Name == "city" && child.Value == "london"));

Here's an example using LINQ to XML.

//load the document
var document = XDocument.Load(@"C:\Development\Website\dvds.xml");
//get all dvd nodes
var dvds = document.Descendants().Where(node => node.Name == "dvd");
//get all dvd nodes that have a city node with a value of "london"
var londonDVDs = dvds.Where(dvd => dvd.Descendants().Any(child => child.Name == "city" && child.Value == "london"));

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