C# 基于子节点对整个xdocument进行排序

我有以下格式的xml：

<?xml version="1.0" encoding="utf-8"?>
<contactGrp name="People">
  <contactGrp name="Developers">
    <customer name="Mike" ></customer>
    <customer name="Brad" ></customer>
    <customer name="Smith" ></customer>
  </contactGrp>
  <contactGrp name="QA">
    <customer name="John" ></customer>
    <customer name="abi" ></customer>
  </contactGrp>
</contactGrp>

<?xml version="1.0" encoding="utf-8"?>
<contactGrp name="People">
  <contactGrp name="Developers">
    <customer name="Brad" ></customer>
    <customer name="Mike" ></customer>
    <customer name="Smith" ></customer>
  </contactGrp>
  <contactGrp name="QA">
    <customer name="abi" ></customer>
    <customer name="John" ></customer>
  </contactGrp>
</contactGrp>

我想根据客户的姓名对客户列表进行排序，并以以下格式返回文档：

<?xml version="1.0" encoding="utf-8"?>
<contactGrp name="People">
  <contactGrp name="Developers">
    <customer name="Mike" ></customer>
    <customer name="Brad" ></customer>
    <customer name="Smith" ></customer>
  </contactGrp>
  <contactGrp name="QA">
    <customer name="John" ></customer>
    <customer name="abi" ></customer>
  </contactGrp>
</contactGrp>

<?xml version="1.0" encoding="utf-8"?>
<contactGrp name="People">
  <contactGrp name="Developers">
    <customer name="Brad" ></customer>
    <customer name="Mike" ></customer>
    <customer name="Smith" ></customer>
  </contactGrp>
  <contactGrp name="QA">
    <customer name="abi" ></customer>
    <customer name="John" ></customer>
  </contactGrp>
</contactGrp>

我正在使用c#，目前使用xmldocument

---

谢谢你

你可以这样做

var doc = XDocument.Load(/* ... */);

foreach (var g in doc.Descendants("contactGrp"))
{
    var customers = g.Elements("customer").ToList();
    customers.Remove();
    g.Add(customers.OrderBy(c => c.Attribute("name").Value));
}

---

如果希望拥有样式表并使用它转换文档，则：

<?xml version="1.0" encoding="utf-8"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
    xmlns:msxsl="urn:schemas-microsoft-com:xslt" exclude-result-prefixes="msxsl"
>
    <xsl:output method="xml" indent="yes"/>

  <xsl:template match="/contactGrp">
    <contactGrp name="Developers">
      <xsl:apply-templates select="contactGrp"/>
    </contactGrp>
  </xsl:template>

  <xsl:template match="contactGrp/contactGrp">
    <contactGrp>
      <xsl:attribute name="name">
        <xsl:value-of select="@name"/>
      </xsl:attribute>

      <xsl:for-each select="customer">
        <xsl:sort select="@name"/>
        <xsl:copy-of select="."/>
      </xsl:for-each>

    </contactGrp>
  </xsl:template>

</xsl:stylesheet>

---

此转换：

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

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

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

应用于提供的XML文档时：

<contactGrp name="People">
    <contactGrp name="Developers">
        <customer name="Mike" ></customer>
        <customer name="Brad" ></customer>
        <customer name="Smith" ></customer>
    </contactGrp>
    <contactGrp name="QA">
        <customer name="John" ></customer>
        <customer name="abi" ></customer>
    </contactGrp>
</contactGrp>

<contactGrp name="People">
  <contactGrp name="Developers">
    <customer name="Brad" />
    <customer name="Mike" />
    <customer name="Smith" />
  </contactGrp>
  <contactGrp name="QA">
    <customer name="abi" />
    <customer name="John" />
  </contactGrp>
</contactGrp>

生成所需的正确结果：

<contactGrp name="People">
    <contactGrp name="Developers">
        <customer name="Mike" ></customer>
        <customer name="Brad" ></customer>
        <customer name="Smith" ></customer>
    </contactGrp>
    <contactGrp name="QA">
        <customer name="John" ></customer>
        <customer name="abi" ></customer>
    </contactGrp>
</contactGrp>

<contactGrp name="People">
  <contactGrp name="Developers">
    <customer name="Brad" />
    <customer name="Mike" />
    <customer name="Smith" />
  </contactGrp>
  <contactGrp name="QA">
    <customer name="abi" />
    <customer name="John" />
  </contactGrp>
</contactGrp>

---

注意事项：无论

contactGrp

元素的嵌套级别如何，始终会生成正确的结果

我感到困惑，因为

xpath

标记。。。如果您想选择一个节点集，那么XPath可以选择正确的技术。如果要对节点集进行正确排序，则需要XPath引擎主机语言。但是如果你想转换一个XML树，标准资源是XSLT。我知道，我似乎无法让它解决我的问题。好问题，+1。请参阅我的答案，以获得一个完整、简短且简单的XSLT解决方案，该解决方案适用于

contactGrp

元素的任何嵌套级别非常感谢。我知道xslt是正确的方法，但无法让xml排序到正确的级别。非常感谢。