Question

我正在用c＃创建一个模型，我的想法是稍后我正在做的应用程序将从xml读取数据并将其转换为对象，但是我想生成一个示例xml。我怎么能这样做？

我的样本c＃类是：

 public class SiteDefinition
    {
        public string Name { get; set; }
        public string Version { get; set; }
        public List<MasterPage> MasterPages { get; set; }
        public List<File> Files { get; set; }
        public List<PageLayout> PageLayouts { get; set; }
        public List<Feature> Features { get; set; }
        public List<ContentType> ContentTypes { get; set; }
        public List<StyleSheet> StyleSheets { get; set; }
    }

Answer 1

var instance = new SiteDefinition();
var serializer = new XmlSerializer(typeof(SiteDefinition));

using(var writer = new StreamWriter("C:\\Path\\To\\File.xml"))
{
    serializer.Serialize(writer, instance);
}

如果你想自定义输出（属性等），你可以使用许多属性来装饰你的类和类成员。查看this MSDN article for more info

Answer 2

XML序列化

必需的命名空间

using System.Xml.Serialization;

以XML格式读写类

    public static List<SiteDefinition> Read()
    {
        XmlSerializer reader = new XmlSerializer(typeof(List<SiteDefinition>));
        using (FileStream file = File.OpenRead(Path.Combine(Global.AppRoot, configFileName)))
        {
            return reader.Deserialize(file) as List<SiteDefinition>;
        }
    }

    public static void Write(List<SiteDefinition> settings)
    {
        XmlSerializer writer = new XmlSerializer(typeof(List<SiteDefinition>));
        using (FileStream file = File.Create(Path.Combine(Global.AppRoot, configFileName)))
        {
            writer.Serialize(file, settings);
        }
    }

如何从类文件生成示例xml

2 个答案: