如果子节点元素相等，如何添加父节点？

Question

我有大量的 xml 数据。因此，"SECTION" 包含 "Piece" 节点，它们对于 "SECTION name="RF1WB-1"、SECTION name="RF1WB-2"、SECTION name="RF1WB-3" 具有相同的值。因此，如果 "Piece" 标记包含与第二个 XML 中所示相同的值，我想将 "Parent" 添加为它们的父节点。

如果子节点元素包含相同的值，如何添加父节点？

XmlDocument doc = new XmlDocument();
doc.Load("..\\MFAB1.xml");        
XmlNodeList xnList = doc.SelectNodes("MFAB.ini/SECTION");
foreach (XmlNode xn in xnList)
{
    if (xn.HasChildNodes)
    {
        foreach (XmlNode item in xn.SelectNodes("Piece"))
        {
            if (xn.ChildNodes[0].InnerText.ToString().Contains(s3))
            {
                d1.Piece = xn.ChildNodes[0].InnerText.ToString();
            }
            ______________
            ______________
            ______________
        }
    }
}

---

输入文件：

<?xml version="1.0" encoding="UTF-8"?>
<MFAB.ini>
 <SECTION name="RF1WB-1">
    <Piece>
      RF1-1
    </Piece>
</SECTION>
  <SECTION name="RF1WB-2">
    <Piece>
      RF1-1
    </Piece>
</SECTION>
  <SECTION name="RF1WB-3">
    <Piece>
      RF1-1
    </Piece>
  </SECTION>
  <SECTION name="RF1-2WB-1">
    <Piece>
      RF1-2
    </Piece>
 </SECTION>
  <SECTION name="RF1-2WB-2">
    <Piece>
      RF1-2
    </Piece>
 </SECTION>
</MFAB.ini>

---

预期的输出文件：

我需要的是：

<?xml version="1.0" encoding="UTF-8"?>
<MFAB.ini>
<Parent name = "RF1-1">
 <SECTION name="RF1-1WB-1">
    <Piece>
      RF1-1
    </Piece>
</SECTION>
  <SECTION name="RF1-1WB-2">
    <Piece>
      RF1-1
    </Piece>
</SECTION>
  <SECTION name="RF1-1WB-3">
    <Piece>
      RF1-1
    </Piece>
  </SECTION>
</Parent>
<Parent name = "RF1-2">
  <SECTION name="RF1-2WB-1">
    <Piece>
      RF1-2
    </Piece>
 </SECTION>
  <SECTION name="RF1-2WB-2">
    <Piece>
      RF1-2
    </Piece>
 </SECTION>
</Parent>
</MFAB.ini>

Answer 1

您可以通过<Piece>标签对元素进行分组，然后将它们作为<Parent>标签的子元素：

var xml = XDocument.Load(@"input file");

var result = new XElement(xml.Root.Name, xml.Root.Attributes().ToArray(),
    xml.Root.Elements().GroupBy(x => x.Element("Piece").Value.Trim()).Select(x =>
        new XElement("Parent", new XAttribute("name", x.Key), x))
    );

result.Save(@"output file");

Answer 2

为了方便您自己，您可以使用XmlSerializer 来读取您的数据，并且您可以像这样为您的 Xml 定义一个 dto

[XmlRoot("MFAB.ini")]
public class Mfab {
    [XmlElement("SECTION")]
    public Section[] Sections { get; set; }
}

public class Section {
    [XmlAttribute("name")]
    public string Name { get; set; }
    [XmlElement("Piece")]
    [XmlText]
    public string Piece { get; set; }
}

这将按原样读取您的文档，将您的所有部分作为 Mfab 类的一部分（抱歉，我不知道这些类的更好名称；））

这个 Xml，你可以这样阅读

private static T GetXmlFromFile<T>(string filename) {
    XmlSerializer xs = new XmlSerializer(typeof(T));
    using (var fs = new FileStream(filename, FileMode.Open, FileAccess.Read))
    {
        return (T)xs.Deserialize(fs);
    }
}

这会给你一个包含所有部分的类。现在，既然你想重新格式化你的数据，你应该创建一个新的类来保存父部分，然后创建一个新的 xml 根来保存它们

[XmlRoot("MFAB.ini")]
public class MfabWithParents {
    [XmlElement("PARENT")]
    public ParentSection[] Parents { get; set; }
}

public class ParentSection {
    [XmlAttribute("name")]
    public string Name { get; set; }
    [XmlElement("SECTION")]
    public Section[] Sections { get; set; }
}

ParentSection 仍然重用之前创建的 Section 类。

现在要进行转换，我们可以执行以下步骤

// get the xml from a file
var source = GetXmlFromFile<Mfab>(Path.Combine(Environment.CurrentDirectory, "data.xml"));
// group all the sections by it's Piece
var groupedSections = source.Sections.GroupBy(section => section.Piece);
// convert the sections from the Grouped section to a parent section with name as attribute
var parentSections = groupedSections.Select(grp => new ParentSection
{
    Name = grp.Key,
    Sections = grp.ToArray()
}).ToArray();
// save the xml to a new file (which could now be read with the MfabWithParents class)
SaveToXml(new MfabWithParents { Parents = parentSections }, Path.Combine(Environment.CurrentDirectory, "data2.xml"));

然后保存方法如下所示

private static void SaveToXml(object data, string filename) {
    XmlSerializer xs = new XmlSerializer(data.GetType());
    using (var fs = new FileStream(filename, FileMode.Create, FileAccess.Write)) {
        xs.Serialize(fs, data);
    }
}