序列化&将属性反序列化为 list<string>使用 C# 编写 xml 文件 [英] serialize & deserialize with property as list<string> xml file using C#

问题描述

如何序列化 &使用 C# 反序列化以下 xml 文件.我已经为这个 xml 创建了可序列化的类.

How to serialize & deserialize below xml file using C#. I have created serializable class for this xml.

在反序列化这个 xml 的一些代码下面，列表只能获得单个值.

below some code to deserialize this xml, the list is able to get only single value.

<?xml version="1.0" encoding="utf-8" ?>
<Configuration>
<CSVFile>
<string>ff</string>
<string>gg</string>
<string>jj</string>
</CSVFile> 
</Configuration>


[Serializable, XmlRoot("Configuration"), XmlType("Configuration")]
public class Configuration
{
    public Configuration()
    {
        CSVFile = new List<string>();
    }

    [XmlElement("CSVFile")]
    public List<string> CSVFile { get; set; }
}

public class Mytutorial
{
    string configFilePath = "above xml file path"

    XmlSerializer serializer = new XmlSerializer(typeof(Configuration));
    FileStream xmlFile = new FileStream(configFilePath, FileMode.Open);
    Configuration con = (Configuration)serializer.Deserialize(xmlFile);
 }

推荐答案

您的 XML 定义与您的模型不匹配.

Your XML definition does not match your models.

<?xml version="1.0" encoding="utf-8" ?>
<Configuration>
  <CSVFile>
    <csvstrings>ff</csvstrings>
    <csvstrings>gg</csvstrings>
    <csvstrings>jj</csvstrings>
  </CSVFile> 
</Configuration>

它需要以下模型:

配置
CSV文件

Configuration
CSVFile

因此，您的实现应该是:

So, your implementation should be:

[Serializable]
public class CSVFile
{
    [XmlElement("csvstrings")]
    public List<string> csvstrings { get; set; }

    public CSVFile()
    {

    }
}

[Serializable, XmlRoot("Configuration"), XmlType("Configuration")]
public class Configuration
{
    public Configuration()
    {

    }

    [XmlElement("CSVFile")]
    public CSVFile csvs { get; set; }
}

public class Mytutorial
{
    string configFilePath = "above xml file path"

    XmlSerializer serializer = new XmlSerializer(typeof(Configuration));
    FileStream xmlFile = new FileStream(configFilePath, FileMode.Open);
    Configuration con = (Configuration)serializer.Deserialize(xmlFile);
}

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