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c# - C#，忽略 XML 反序列化的大小写敏感

问题描述

1.xml文件包含xml元素：“INSTITUTION”和“institution”，每个xml元素都包含名称、代码和INN等属性。

如果 xml 元素名称为：“INSTITUTION”，我的代码可以正常工作，并且在到达 xml 元素名称：“institution”时会失败。

我从stackoverflow中找到了一些解决方案，但对我没有帮助......

如何忽略 xml 反序列化的大小写敏感？ 

<institutions>
<INSTITUTION name= "some_value" CODE="some_value" INN="some_value"/>
<INSTITUTION name= "some_value" CODE="some_value" INN="some_value"/>
<INSTITUTION name= "some_value" CODE="some_value" INN="some_value"/>
<institution name= "some_value" code="some_value" inn="some_value"/>
<institution name= "some_value" code="some_value" inn="some_value"/>
<institution name= "some_value" code="some_value" inn="some_value"/>
<institution name= "some_value" code="some_value" inn="some_value"/>
</institutions>
</treasury>

C#代码：

using System;
using System.Collections.Generic;
using System.IO;
using System.Text;
using System.Xml;
using System.Xml.Serialization;
using Test_for_finding_file_type.XmlSynonymDeserializer;
//This is the class that will be deserialized
[XmlRoot("treasury")]
public class Treasury
{
    [XmlElement("institutions")]
    public institutions Institutions { get; set; }
}

public class institutions
{
    [XmlElement("INSTITUTION")]
    public List<Institution> InstitutionList { get; set; }
}
public class Institution
{
    [XmlAttribute("name")]
    public string Name;
    [XmlAttribute("CODE")]
    public string Code;
    [XmlAttribute("INN")]
    public string Inn;
}


public class Program
{
    public static void Main(String[] args)
    {
        Program pro = new Program();
        pro.DeserializeObject("test.xml");
    }
    private void DeserializeObject(string filename)
    {
        Console.WriteLine("Reading with XML Reader");

        XmlSerializer serializer = new XmlSerializer(typeof(Treasury));


        FileStream fs = new FileStream(filename, FileMode.Open);
        XmlReader reader = XmlReader.Create(fs);


        Treasury treasuryAccounts;
        treasuryAccounts = (Treasury)serializer.Deserialize(reader);

        fs.Close();

        Console.WriteLine("\n------------------------------------------Institutions---------------------------------------------------------\n");
        foreach (var institition in treasuryAccounts.Institutions.InstitutionList)
        {
            Console.Write("Treasury Account Name:" + institition.Name
                  + "\tCODE:" + institition.Code
                  + "\tINN:" + institition.Inn
                  + "\n\n"
          );
        }

        Console.ReadKey();
    }
}

标签： c#xmlxml-deserialization

解决方案

整个问题是缺少 XmlRoot 属性。请参阅下面的代码：

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

[XmlRoot("treasury")]
public class Treasury
{
    [XmlElement("institutions")]
    public Institutions institutions { get; set; }
}
[XmlRoot("institutions")]
public class Institutions
{
    [XmlElement("INSTITUTION")]
    public List<Institution> InstitutionList { get; set; }
}
public class Institution
{
    [XmlAttribute("name")]
    public string Name;
    [XmlAttribute("CODE")]
    public string Code;
    [XmlAttribute("INN")]
    public string Inn;
}


public class Program
{
    const string FILENAME = @"c:\temp\test.xml";
    public static void Main(String[] args)
    {
        Program pro = new Program();
        pro.DeserializeObject(FILENAME);
    }
    private void DeserializeObject(string filename)
    {
        Console.WriteLine("Reading with XML Reader");

        XmlSerializer serializer = new XmlSerializer(typeof(Institutions));

        XmlReader reader = XmlReader.Create(FILENAME);


        Treasury treasuryAccounts = new Treasury();

        treasuryAccounts.institutions = (Institutions)serializer.Deserialize(reader);

        reader.Close();

        Console.WriteLine("\n------------------------------------------Institutions---------------------------------------------------------\n");
        foreach (var institition in treasuryAccounts.institutions.InstitutionList)
        {
            Console.Write("Treasury Account Name:" + institition.Name
                  + "\tCODE:" + institition.Code
                  + "\tINN:" + institition.Inn
                  + "\n\n"
          );
        }

        Console.ReadKey();
    }
}

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