我尝试在一个 XML 文件中保存和读取多个对象。
序列化功能不适用于我现有的列表,但我不知道为什么。我已经尝试编译它,但收到一个错误,指出该方法需要对象引用。
程序.cs:
class Program
{
static void Main(string[] args)
{
List<Cocktail> lstCocktails = new List<Cocktail>();
listCocktails.AddRange(new Cocktail[]
{
new Cocktail(1,"Test",true,true,
new Cocktail(1, "Test4", true, true, 0)
});
Serialize(lstCocktails);
}
public void Serialize(List<Cocktail> list)
{
XmlSerializer serializer = new XmlSerializer(typeof(List<Cocktail>));
using (TextWriter writer = new StreamWriter(@"C:\Users\user\Desktop\MapSample\bin\Debug\ListCocktail.xml"))
{
serializer.Serialize(writer, list);
}
}
private void DiserializeFunc()
{
var myDeserializer = new XmlSerializer(typeof(List<Cocktail>));
using (var myFileStream = new FileStream(@"C:\Users\user\Desktop\MapSample\bin\Debug\ListCocktail.xml", FileMode.Open))
{
ListCocktails = (List<Cocktail>)myDeserializer.Deserialize(myFileStream);
}
}
鸡尾酒.cs:
[Serializable()]
[XmlRoot("locations")]
public class Cocktail
{
[XmlElement("id")]
public int CocktailID { get; set; }
[XmlElement("name")]
public string CocktailName { get; set; }
[XmlElement("alc")]
public bool alcohol { get; set; }
[XmlElement("visible")]
public bool is_visible { get; set; }
[XmlElement("counter")]
public int counter { get; set; }
private XmlSerializer ser;
public Cocktail() {
ser = new XmlSerializer(this.GetType());
}
public Cocktail(int id, string name, bool alc,bool vis,int count)
{
this.CocktailID = id;
this.CocktailName = name;
this.alcohol = alc;
this.is_visible = vis;
this.counter = count;
}
}
}
我还认为我把 DiserializeFunc() 搞砸了。
您非常接近正确实现 Cocktail 类,但我认为您对如何序列化列表感到困惑。您对 Cocktail 对象类的实现完全没问题,只需去掉列表相关函数即可。
using System;
using System.Xml.Serialization;
namespace Serialization_Help
{
[Serializable()]
[XmlRoot("locations")]
public class Cocktail
{
[XmlElement("id")]
public int CocktailID { get; set; }
[XmlElement("name")]
public string CocktailName { get; set; }
[XmlElement("alc")]
public bool alcohol { get; set; }
[XmlElement("visible")]
public bool is_visible { get; set; }
[XmlElement("counter")]
public int counter { get; set; }
public Cocktail() {
}
public Cocktail(int id, string name, bool alc, bool vis, int count)
{
this.CocktailID = id;
this.CocktailName = name;
this.alcohol = alc;
this.is_visible = vis;
this.counter = count;
}
}
}
现在在新函数中,您想直接序列化列表。
using System.Collections.Generic;
using System.IO;
using System.Xml.Serialization;
namespace Serialization_Help
{
class Program {
static void Main(string[] args) {
List<Cocktail> list = new List<Cocktail> {
new Cocktail(01, "rum and coke", true, true, 5),
new Cocktail(02, "water on the rocks", false, true, 3)
};
Serialize(list);
List<Cocktail> deserialized = DiserializeFunc();
}
public static void Serialize(List<Cocktail> list) {
XmlSerializer serializer = new XmlSerializer(typeof(List<Cocktail>));
using (TextWriter writer = new StreamWriter(Directory.GetCurrentDirectory() + @"\ListCocktail.xml")) serializer.Serialize(writer, list);
}
private static List<Cocktail> DiserializeFunc() {
var myDeserializer = new XmlSerializer(typeof(List<Cocktail>));
using (var myFileStream = new FileStream(Directory.GetCurrentDirectory() + @"\ListCocktail.xml", FileMode.Open)) return (List<Cocktail>)myDeserializer.Deserialize(myFileStream);
}
}
}
这样做应该正确打印出以下 .xml 输出:
<?xml version="1.0" encoding="utf-8"?>
<ArrayOfCocktail xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema">
<Cocktail>
<id>1</id>
<name>rum and coke</name>
<alc>true</alc>
<visible>true</visible>
<counter>5</counter>
</Cocktail>
<Cocktail>
<id>2</id>
<name>water on the rocks</name>
<alc>false</alc>
<visible>true</visible>
<counter>3</counter>
</Cocktail>
</ArrayOfCocktail>
请记住,我没有提供文件的任何标准安全或空检查的实现。您必须使用
File.Exists(...)
(请参阅此处了解 File.Exists 实现) 自行检查文件是否存在,并实现正确的 try
和 catch
案例以及您的代码在运行时将选择执行的操作导致序列化或输入/输出错误。
你最好使用ExtendedXmlSerializer来序列化和反序列化。
安装 您可以从 nuget 安装 ExtendedXmlSerializer 或运行以下命令:
Install-Package ExtendedXmlSerializer
连载:
ExtendedXmlSerializer serializer = new ExtendedXmlSerializer();
var list = new List<Cocktail>();
var xml = serializer.Serialize(list);
反序列化
var list = serializer.Deserialize<List<Cocktail>>(xml);
.NET 中的标准 XML 序列化程序非常有限。
ExtendedXmlSerializer 可以做到这一点以及更多。
ExtendedXmlSerializer 支持 .NET 4.5 或更高版本以及 .NET Core。您可以将其与 WebApi 和 AspCore 集成。