C#XML序列化非嵌套数组/列表

我正在编写一个C#

WinForm

应用程序，它应该读取一个预先存在的

XML

文件。 我需要解析

XML

文件，并建立一个数据结构来反映

XML

文件内容。 我在

XML

序列化方面有一些经验，因此我决定使用

.NET

XML.serialization

功能。 我陷入了一个奇怪的

XML

结构中，我无法在类中匹配它（通过属性、元素等）：

我还有一个次要问题。反映此结构的最佳类层次结构是： a） 以与xml元素相同的方式嵌套类 b） 将所有类保持在同一级别（没有层次结构）？

---

有一些副作用需要考虑？

您可以如下定义

C

对象

[XmlRoot(ElementName="comment")]
public class Comment {
    [XmlAttribute(AttributeName="number")]
    public string Number { get; set; }
    [XmlAttribute(AttributeName="value")]
    public string Value { get; set; }
}

[XmlRoot(ElementName="title_block")]
public class Title_block {
    [XmlElement(ElementName="title")]
    public string Title { get; set; }
    [XmlElement(ElementName="company")]
    public string Company { get; set; }
    [XmlElement(ElementName="rev")]
    public string Rev { get; set; }
    [XmlElement(ElementName="date")]
    public string Date { get; set; }
    [XmlElement(ElementName="source")]
    public string Source { get; set; }
    [XmlElement(ElementName="comment")]
    public List<Comment> Comment { get; set; }
}

[XmlRoot(ElementName="sheet")]
public class Sheet {
    [XmlElement(ElementName="title_block")]
    public Title_block Title_block { get; set; }
    [XmlAttribute(AttributeName="number")]
    public string Number { get; set; }
    [XmlAttribute(AttributeName="name")]
    public string Name { get; set; }
    [XmlAttribute(AttributeName="tstamps")]
    public string Tstamps { get; set; }
}

---

选中此项

您可以按如下方式定义

C

对象

[XmlRoot(ElementName="comment")]
public class Comment {
    [XmlAttribute(AttributeName="number")]
    public string Number { get; set; }
    [XmlAttribute(AttributeName="value")]
    public string Value { get; set; }
}

[XmlRoot(ElementName="title_block")]
public class Title_block {
    [XmlElement(ElementName="title")]
    public string Title { get; set; }
    [XmlElement(ElementName="company")]
    public string Company { get; set; }
    [XmlElement(ElementName="rev")]
    public string Rev { get; set; }
    [XmlElement(ElementName="date")]
    public string Date { get; set; }
    [XmlElement(ElementName="source")]
    public string Source { get; set; }
    [XmlElement(ElementName="comment")]
    public List<Comment> Comment { get; set; }
}

[XmlRoot(ElementName="sheet")]
public class Sheet {
    [XmlElement(ElementName="title_block")]
    public Title_block Title_block { get; set; }
    [XmlAttribute(AttributeName="number")]
    public string Number { get; set; }
    [XmlAttribute(AttributeName="name")]
    public string Name { get; set; }
    [XmlAttribute(AttributeName="tstamps")]
    public string Tstamps { get; set; }
}

检查此项

尝试此项

使用

using System.Collections.Generic;
using System.IO;
using System.Text;
using System.Xml;
using System.Xml.Serialization;

班级

[XmlRoot(ElementName = "comment")]
public class Comment
{
    [XmlAttribute(AttributeName = "number")]
    public string Number { get; set; }
    [XmlAttribute(AttributeName = "value")]
    public string Value { get; set; }
}

[XmlRoot(ElementName = "title_block")]
public class Title_block
{
    [XmlElement(ElementName = "title")]
    public string Title { get; set; }
    [XmlElement(ElementName = "company")]
    public string Company { get; set; }
    [XmlElement(ElementName = "rev")]
    public string Rev { get; set; }
    [XmlElement(ElementName = "date")]
    public string Date { get; set; }
    [XmlElement(ElementName = "source")]
    public string Source { get; set; }
    [XmlElement(ElementName = "comment")]
    public List<Comment> Comment { get; set; }
}

[XmlRoot(ElementName = "sheet")]
public class Sheet
{
    [XmlElement(ElementName = "title_block")]
    public Title_block Title_block { get; set; }
    [XmlAttribute(AttributeName = "number")]
    public string Number { get; set; }
    [XmlAttribute(AttributeName = "name")]
    public string Name { get; set; }
    [XmlAttribute(AttributeName = "tstamps")]
    public string Tstamps { get; set; }
}

这段代码从一个文件（在application*.exe文件夹中称为xml.xml）中读取xml，然后将其重新序列化为一个名为dezerializedXML的对象

试试这个

使用

using System.Collections.Generic;
using System.IO;
using System.Text;
using System.Xml;
using System.Xml.Serialization;

班级

[XmlRoot(ElementName = "comment")]
public class Comment
{
    [XmlAttribute(AttributeName = "number")]
    public string Number { get; set; }
    [XmlAttribute(AttributeName = "value")]
    public string Value { get; set; }
}

[XmlRoot(ElementName = "title_block")]
public class Title_block
{
    [XmlElement(ElementName = "title")]
    public string Title { get; set; }
    [XmlElement(ElementName = "company")]
    public string Company { get; set; }
    [XmlElement(ElementName = "rev")]
    public string Rev { get; set; }
    [XmlElement(ElementName = "date")]
    public string Date { get; set; }
    [XmlElement(ElementName = "source")]
    public string Source { get; set; }
    [XmlElement(ElementName = "comment")]
    public List<Comment> Comment { get; set; }
}

[XmlRoot(ElementName = "sheet")]
public class Sheet
{
    [XmlElement(ElementName = "title_block")]
    public Title_block Title_block { get; set; }
    [XmlAttribute(AttributeName = "number")]
    public string Number { get; set; }
    [XmlAttribute(AttributeName = "name")]
    public string Name { get; set; }
    [XmlAttribute(AttributeName = "tstamps")]
    public string Tstamps { get; set; }
}

---

这段代码从一个文件（在application*.exe文件夹中称为xml.xml）中读取xml，然后将其重新序列化为一个名为dezerializedXML的对象

我们希望你的问题已经解决。如果没有，也许可以考虑更新你的问题。。。。我们出去度假了，我马上去看看。我们希望你的问题得到解决。如果没有，也许可以考虑更新你的问题。。。。我出去度假了，我很快就去看看

try
{
    XmlDocument xmlDoc = new XmlDocument();
    xmlDoc.Load("xml.xml");

    string XML = xmlDoc.InnerXml.ToString();
    byte[] BUFXML = ASCIIEncoding.UTF8.GetBytes(XML);
    MemoryStream ms1 = new MemoryStream(BUFXML);

    XmlSerializer DeserializerPlaces = new XmlSerializer(typeof(Sheet));
    using (XmlReader reader = new XmlTextReader(ms1))
    {
        Sheet dezerializedXML = (Sheet)DeserializerPlaces.Deserialize(reader);

    }// Put a break-point here, then mouse-over dezerializedXML and you should have you values
}
catch (System.Exception)
{
    throw;
}