本文介绍了从xml文件映射到类属性的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
Hi I have this xml file, and I want to deserialize this xml file to a class which has exactly the same properties like this xml file, I know serialization and deserialization, but I don't know how to map the properties exactly!!!
<?xml version="1.0" encoding="utf-8" ?>
- <XML_FILE_FOR_SOME_INPUTS>
- <Printer1>
<Title>behnoud</Title>
<PhoneNumber>22174960</PhoneNumber>
<Site>www.sbu.com</Site>
<PrinterPort>abc</PrinterPort>
</Printer1>
- <Printer2>
<BillAcceptorPort>abc</BillAcceptorPort>
</Printer2>
</XML_FILE_FOR_SOME_INPUTS>
推荐答案
public class InputToolsClass
{
public InputToolsClass()
{
}
public InputToolsClass(string companytitle, string phonenumber, string site, string printer1, string Printer2 )
{
this.Title = title;
this.PhoneNumber = phonenumber;
this.Site = site;
this.Printer1 = printer1;
this.Printer2 = Printer2 ;
}
public string Title { get; set; }
public string PhoneNumber { get; set; }
public string Site { get; set; }
public string Printer1 { get; set; }
public string Printer2 { get; set; }
}
public void SetAllProperties()
{
XmlSerializer ser = new XmlSerializer(typeof(InputToolsClass));
InputToolsClass inputs;
using (StreamReader reader = new StreamReader(@"C:\Desktop\Inputs.xml"))
{
inputs = (InputToolsClass)ser.Deserialize(reader);
}
}
XmlRootAttribute root = new XmlRootAttribute("XML_FILE_FOR_SOME_INPUTS");
因此必须添加XmlRootAttribute。
alose it is essential to add XmlRootAttribute.
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