C# 如何将 List<T> 序列化为 XML？
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How to serialize a List<T> into XML?
提问by Ujjwal27
How to convert this list:
如何转换此列表：
List<int> Branches = new List<int>();
Branches.Add(1);
Branches.Add(2);
Branches.Add(3);
into this XML:
进入这个 XML：
<Branches>
<branch id="1" />
<branch id="2" />
<branch id="3" />
</Branches>
采纳答案by Praveen
You can try this using LINQ:
您可以使用 LINQ 尝试此操作：
List<int> Branches = new List<int>();
Branches.Add(1);
Branches.Add(2);
Branches.Add(3);
XElement xmlElements = new XElement("Branches", Branches.Select(i => new XElement("branch", i)));
System.Console.Write(xmlElements);
System.Console.Read();
Output:
输出：
<Branches>
<branch>1</branch>
<branch>2</branch>
<branch>3</branch>
</Branches>
Forgot to mention: you need to include using System.Xml.Linq;namespace.
忘了提：你需要包含using System.Xml.Linq;命名空间。
EDIT:
编辑：
XElement xmlElements = new XElement("Branches", Branches.Select(i => new XElement("branch", new XAttribute("id", i))));
XElement xmlElements = new XElement("Branches", Branches.Select(i => new XElement("branch", new XAttribute("id", i))));
output:
输出：
<Branches>
<branch id="1" />
<branch id="2" />
<branch id="3" />
</Branches>
回答by Dustin Kingen
You can use Linq-to-XML
您可以使用Linq-to-XML
List<int> Branches = new List<int>();
Branches.Add(1);
Branches.Add(2);
Branches.Add(3);
var branchesXml = Branches.Select(i => new XElement("branch",
new XAttribute("id", i)));
var bodyXml = new XElement("Branches", branchesXml);
System.Console.Write(bodyXml);
Or create the appropriate class structure and use XML Serialization.
或者创建适当的类结构并使用XML 序列化。
[XmlType(Name = "branch")]
public class Branch
{
[XmlAttribute(Name = "id")]
public int Id { get; set; }
}
var branches = new List<Branch>();
branches.Add(new Branch { Id = 1 });
branches.Add(new Branch { Id = 2 });
branches.Add(new Branch { Id = 3 });
// Define the root element to avoid ArrayOfBranch
var serializer = new XmlSerializer(typeof(List<Branch>),
new XmlRootAttribute("Branches"));
using(var stream = new StringWriter())
{
serializer.Serialize(stream, branches);
System.Console.Write(stream.ToString());
}
