以下代码给我报错:
我需要怎么改变才能使用 Foreach 遍历我的'System.Xml.Linq.XElement.XElement(System.Xml.Linq.XName,object)'的最佳重载方法匹配存在一些无效参数。
List<Customer>
集合并构建 XDocument?
using System;
using System.Collections.Generic;
using System.Xml.Linq;
namespace test_xml3
{
class Program
{
static void Main(string[] args)
{
List<Customer> customers = new List<Customer> {
new Customer {FirstName="Jim", LastName="Smith", Age=27},
new Customer {FirstName="Hank", LastName="Moore", Age=28},
new Customer {FirstName="Jay", LastName="Smythe", Age=44}
};
Console.WriteLine(BuildXmlWithLINQ(customers));
Console.ReadLine();
}
private static string BuildXmlWithLINQ(List<Customer> customers)
{
XDocument xdoc = new XDocument
(
new XDeclaration("1.0", "utf-8", "yes"),
new XElement("customers",
customers.ForEach(c => new XElement("customer",
new XElement("firstName", c.FirstName),
new XElement("lastName", c.LastName)
)
)
);
return xdoc.Declaration.ToString() + Environment.NewLine + xdoc.ToString();
}
}
public class Customer
{
public string FirstName { get; set; }
public string LastName { get; set; }
public int Age { get; set; }
}
}
新增:
感谢您的所有答案,我也想出了这个可以运作的方法:
private static string BuildXmlWithLINQ2(List<Customer> customers)
{
XDocument xdoc = new XDocument(
new XDeclaration("1.0", "utf-8", "yes")
);
XElement xRoot = new XElement("customers");
xdoc.Add(xRoot);
foreach (var customer in customers)
{
XElement xElement = new XElement("customer",
new XElement("firstName", customer.FirstName),
new XElement("lastName", customer.LastName),
new XElement("age", customer.Age)
);
xRoot.Add(xElement);
}
return xdoc.Declaration.ToString() + Environment.NewLine + xdoc.ToString();
}
... c=> new XElement(...
周围加上括号。 - ULysses