[英]C#, Ignore case sensitivity for XML deserialization
1.Xml 文件包含 xml 元素:“机构”和“机构”,每个 xml 元素包含名称、代码和 INN 等属性。
如果 xml 元素名称为:“INSTITUTION”,并且在到达 xml 元素名称:“institution”时失败,我的代码工作正常。
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性?
<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();
}
}
整个问题是缺少 XmlRoot 属性。 请参见下面的代码:
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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