怎么快速的抓好List为XML
在本文中，我将演示一个使用LINQ将List <>转换为XML的简单示例。我假设你熟悉LINQ to XML。如果不是这样，我建议你亲自动手学习LINQ to XML。这三篇文章将有助于：
使用LINQ to XML的一些常见操作 - 第一部分
使用LINQ to XML的一些常见操作 - 第二部分
使用LINQ to XML的一些常见操作 - 第III部分
让我们开始吧。第一步是定义List <>。出于演示目的，我将创建一个控制台应用程序并声明一个List <> of Employees，如下所示：
C#
using System;
using System.Collections.Generic;
using System.Linq;
using System.Xml.Linq;
namespace ListToXML
{
class Program
{
static void Main(string[] args)
{
List<Employee> empList = new List<Employee>();
empList.Add(new Employee() { ID = 1, FName = "John", LName = "Shields", DOB =DateTime.Parse("12/11/1971"), Sex = 'M' });
empList.Add(new Employee() { ID = 2, FName = "Mary", LName = "Jacobs", DOB =DateTime.Parse("01/17/1961"), Sex = 'F' });
empList.Add(new Employee() { ID = 3, FName = "Amber", LName = "Agar", DOB =DateTime.Parse("12/23/1971"), Sex = 'M' });
empList.Add(new Employee() { ID = 4, FName = "Kathy", LName = "Berry", DOB =DateTime.Parse("11/15/1976"), Sex = 'F' });
empList.Add(new Employee() { ID = 5, FName = "Lena", LName = "Bilton", DOB =DateTime.Parse("05/11/1978"), Sex = 'F' });
empList.Add(new Employee() { ID = 6, FName = "Susanne", LName = "Buck", DOB =DateTime.Parse("03/7/1965"), Sex = 'F' });
empList.Add(new Employee() { ID = 7, FName = "Jim", LName = "Brown", DOB =DateTime.Parse("09/11/1972"), Sex = 'M' });
empList.Add(new Employee() { ID = 8, FName = "Jane", LName = "Hooks", DOB =DateTime.Parse("12/11/1972"), Sex = 'F' });
empList.Add(new Employee() { ID = 9, FName = "Robert", LName = "", DOB =DateTime.Parse("06/28/1964"), Sex = 'M' });
empList.Add(new Employee() { ID = 10, FName = "Cindy", LName = "Fox", DOB =DateTime.Parse("01/11/1978"), Sex = 'M' });
}
}
class Employee
{
public int ID { get; set; }
public string FName { get; set; }
public string LName { get; set; }
public DateTime DOB { get; set; }
public char Sex { get; set; }
}
}
VB.NET
Option Infer On
Imports System
Imports System.Collections.Generic
Imports System.Linq
Imports System.Xml.Linq
Module Module1
Sub Main(ByVal args() As String)
Dim empList As New List(Of Employee)()
empList.Add(New Employee() With {.ID = 1, .FName = "John", .LName = "Shields", .DOB =Date.Parse("12/11/1971"), .Sex = "M"c})
empList.Add(New Employee() With {.ID = 2, .FName = "Mary", .LName = "Jacobs", .DOB =Date.Parse("01/17/1961"), .Sex = "F"c})
empList.Add(New Employee() With {.ID = 3, .FName = "Amber", .LName = "Agar", .DOB =Date.Parse("12/23/1971"), .Sex = "M"c})
empList.Add(New Employee() With {.ID = 4, .FName = "Kathy", .LName = "Berry", .DOB =Date.Parse("11/15/1976"), .Sex = "F"c})
empList.Add(New Employee() With {.ID = 5, .FName = "Lena", .LName = "Bilton", .DOB =Date.Parse("05/11/1978"), .Sex = "F"c})
empList.Add(New Employee() With {.ID = 6, .FName = "Susanne", .LName = "Buck", .DOB =Date.Parse("03/7/1965"), .Sex = "F"c})
empList.Add(New Employee() With {.ID = 7, .FName = "Jim", .LName = "Brown", .DOB =Date.Parse("09/11/1972"), .Sex = "M"c})
empList.Add(New Employee() With {.ID = 8, .FName = "Jane", .LName = "Hooks", .DOB =Date.Parse("12/11/1972"), .Sex = "F"c})
empList.Add(New Employee() With {.ID = 9, .FName = "Robert", .LName = "", .DOB =Date.Parse("06/28/1964"), .Sex = "M"c})
empList.Add(New Employee() With {.ID = 10, .FName = "Cindy", .LName = "Fox", .DOB =Date.Parse("01/11/1978"), .Sex = "M"c})
End Sub
End Module
Friend Class Employee
Private privateID As Integer
Public Property ID() As Integer
Get
Return privateID
End Get
Set(ByVal value As Integer)
privateID = value
End Set
End Property
Private privateFName As String
Public Property FName() As String
Get
Return privateFName
End Get
Set(ByVal value As String)
privateFName = value
End Set
End Property
Private privateLName As String
Public Property LName() As String
Get
Return privateLName
End Get
Set(ByVal value As String)
privateLName = value
End Set
End Property
Private privateDOB As Date
Public Property DOB() As Date
Get
Return privateDOB
End Get
Set(ByVal value As Date)
privateDOB = value
End Set
End Property
Private privateSex As Char
Public Property Sex() As Char
Get
Return privateSex
End Get
Set(ByVal value As Char)
privateSex = value
End Set
End Property
End Class
The next step is to convert this List<> to XML. The root node in our XML, without a doubt, can be <Employees>. Now if you observe, the Employee ID property uniquely identifies each Employee. We can create Parent Nodes with the attribute ID and all the other properties like Fname, Lname, Dob and Age will be the Child Nodes. Here’s the LINQ query to do so:
C#
try
{
var xEle = new XElement("Employees",
from emp in empList
select new XElement("Employee",
new XAttribute("ID", emp.ID),
new XElement("FName", emp.FName),
new XElement("LName", emp.LName),
new XElement("DOB", emp.DOB),
new XElement("Sex", emp.Sex)
));
xEle.Save("D:\\employees.xml");
Console.WriteLine("Converted to XML");
}
catch (Exception ex)
{
Console.WriteLine(ex.Message);
}
Console.ReadLine();
VB.NET
Try
Dim xEle = New XElement("Employees", _
From emp In empList _
Select New XElement("Employee", New XAttribute("ID", emp.ID), _
New XElement("FName", emp.FName), _
New XElement("LName", emp.LName), _
New XElement("DOB", emp.DOB), _
New XElement("Sex", emp.Sex)))
xEle.Save("D:\employees.xml")
Console.WriteLine("Converted to XML")
Catch ex As Exception
Console.WriteLine(ex.Message)
End Try
Console.ReadLine()
在运行应用程序时，输出类似于以下内容：
