我在xml文件下面有这个文件:-
<?xml version="1.0" encoding="utf-8" ?>
<LoanProduct xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance">
<Program>
<ProgramID>6</ProgramID>
<Name>Primary Loan</Name>
<InterestRate>0.23</InterestRate>
<StartDate>2018-12-20</StartDate>
<EndDate>2019-03-31</EndDate>
</Program>
<Program>
<ProgramID>6</ProgramID>
<Name>Primary Loan</Name>
<InterestRate>0.25</InterestRate>
<StartDate>2019-04-1</StartDate>
<EndDate>2099-12-31</EndDate>
</Program>
</LoanProduct>
在我的类文件中,我必须读取xml文件并对其进行一些查询:-
String xml = Server.MapPath("/Resources/LoanProduct.xml");
DataSet dataSet = new DataSet();
dataSet.ReadXml(xml);
假设我要检索ProgramID = 6和EndDate ='2099-12-31'的位置 我该如何实现?
答案 0 :(得分:1)
您可以通过在命名空间XDocument
下使用System.Xml.Linq
来获得所需的结果。
XDocument doc = XDocument.Load(@"Path to your xml file");
var ns = doc.Root.GetDefaultNamespace();
var result = (from program in doc.Descendants(ns + "Program")
where Convert.ToInt32(program.Element(ns + "ProgramID").Value) == 6 && program.Element(ns + "EndDate").Value == "2099-12-31"
select program).FirstOrDefault();
输出:(来自调试器)
答案 1 :(得分:0)
您可以使用序列化。日期应保持一致,否则此代码将导致错误。
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Xml;
using System.Xml.Serialization;
namespace ConsoleApplication3
{
class Program1
{
const string FILENAME = @"c:\temp\test.xml";
static void Main(string[] args)
{
XmlReader reader = XmlReader.Create(FILENAME);
XmlSerializer serializer = new XmlSerializer(typeof(LoanProduct));
LoanProduct loanProduct = (LoanProduct)serializer.Deserialize(reader);
}
}
[XmlRoot(ElementName = "LoanProduct", Namespace = "")]
public class LoanProduct
{
[XmlElement("Program", Namespace = "")]
public List<Program> programs { get; set; }
}
[XmlRoot(ElementName = "Program", Namespace = "")]
public class Program
{
public int ProgramID { get; set; }
public string Name { get; set; }
public decimal InterestRate { get; set; }
public DateTime StartDate { get; set; }
public DateTime EndDate { get; set; }
}
}