如何用C#将XML文件转换为CSV文件

Question

给我最好的方式来写代码。

如何在C#中将XML文件转换为CSV文件，只显示以下标记：

   <a:MarketingAllCardholder xmlns:b="http://schemas.datacontract.org/2004/07/ExternalClient.Data.Classes">
       <b:MarketingAllCardholderData>
          <b:CentreName>Corporate Office</b:CentreName>
          <b:Country>Austria</b:Country>
          <b:CustomerId>379</b:CustomerId>
          <b:DOB>25/02/1991</b:DOB>
          <b:Email>[e-mail]</b:Email>
          <b:ExpiryDate>03/01/2020 08:01</b:ExpiryDate>

       </b:MarketingAllCardholderData>
       <b:MarketingAllCardholderData>
          <b:CentreName>Corporate Office</b:CentreName>
          <b:Country>Egypt</b:Country>
          <b:CustomerId>988915</b:CustomerId>
          <b:DOB>01/03/1986</b:DOB>
          <b:Email>[e-mail]</b:Email>
          <b:ExpiryDate>07/01/2020 11:38</b:ExpiryDate>

       </b:MarketingAllCardholderData>

输出CSV文件格式应如下所示：

Answer 1

基本上可以使用XMLSerializator或XSLT。

使用XmlSerializator：

public class MarketingAllCardholder{
    public MarketingAllCardholderData[] marketingAllCardholderDataList { get; set; }
}

using System.IO;
using System.Xml.Serialization;

using (FileStream fs = new FileStream(@"name.xml", FileMode.Open))
{
    XmlSerializer serializer = new XmlSerializer(typeof(MarketingAllCardholde[]));
    var data = (MarketingAllCardholder[])serializer.Deserialize(fs);
    List<string> list = new List<string>();
    foreach (var item in data)
    {
        //Add All the necessary columns here...
        //After the columns add the delimiter character -> string.Join(","....
    }
    File.WriteAllLines("D:\\csvFile.csv", list);
}

另一种方法是使用XSLT来转换它。

null

public static string ToCSV(string xmlTextDate, string xsltFile)
{
  string result = string.Empty;
  var xpathDoc = new XPathDocument(xmlTextDate);
  var xsltTransform = new System.Xml.Xsl.XslCompiledTransform();
  xsltTransform.Load(xsltFile);

  using (MemoryStream ms = new MemoryStream())
  {
      var writer = new XmlTextWriter(ms, Encoding.UTF8);
      using (var rd = new StreamReader(ms))
      {
          var argList = new System.Xml.Xsl.XsltArgumentList();
          xsltTransform.Transform(xpathDoc, argList, writer);
          ms.Position = 0;
          result = rd.ReadToEnd();
      }
  }
  return result;
}

var csvString = ToCSV("name.xml","name.xsl");