5

我有下表：

CREATE TABLE [dbo].[Data] (
    [Id]            UNIQUEIDENTIFIER NOT NULL,
    [Data]   XML              NOT NULL,
);

我需要将它映射到对象：

class Data
{
    public virtual Guid Id {get; set;}
    public virtual StronglyTypedData Data {get; set;}
}

其中，StronglyTypedData 类似于：

class StronglyTypedData
{
    public string Name {get; set;}
    public int Number {get; set;}
}

默认情况下，XML 列映射到 XmlDocument 属性，但我希望 XML 序列化/反序列化到 StronglyTypedData 属性而不是在映射时发生。

我需要做什么才能做到这一点？

4

2 回答 2

8

我打算对 Diego 的帖子发表评论，但它太长了，我想要语法突出显示。我修改了 Diego 发布的 XmlDocType，以便它使用 xml 序列化来往来于强类型对象。

我制作了自己的通用 IUserType 来处理强类型：

//you'll need these at the top of your file
//using System;
//using System.Collections.Generic;
//using System.Linq;
//using System.Text;
//using NHibernate.UserTypes;
//using NHibernate.SqlTypes;
//using System.Data;
//using System.Xml;
//using NHibernate.Type;

[Serializable]
public class XmlType<T> : MutableType
{
    public XmlType()
        : base(new XmlSqlType())
    {
    }


    public XmlType(SqlType sqlType)
        : base(sqlType)
    {
    }

    public override string Name
    {
        get { return "XmlOfT"; }
    }

    public override System.Type ReturnedClass
    {
        get { return typeof(T); }
    }

    public override void Set(IDbCommand cmd, object value, int index)
    {
        ((IDataParameter)cmd.Parameters[index]).Value = XmlUtil.ConvertToXml(value);
    }

    public override object Get(IDataReader rs, int index)
    {
        // according to documentation, GetValue should return a string, at list for MsSQL
        // hopefully all DataProvider has the same behaviour
        string xmlString = Convert.ToString(rs.GetValue(index));
        return FromStringValue(xmlString);
    }

    public override object Get(IDataReader rs, string name)
    {
        return Get(rs, rs.GetOrdinal(name));
    }

    public override string ToString(object val)
    {
        return val == null ? null : XmlUtil.ConvertToXml(val);
    }

    public override object FromStringValue(string xml)
    {
        if (xml != null)
        {
            return XmlUtil.FromXml<T>(xml);
        }
        return null;
    }

    public override object DeepCopyNotNull(object value)
    {
        var original = (T)value;
        var copy = XmlUtil.FromXml<T>(XmlUtil.ConvertToXml(original));
        return copy;
    }

    public override bool IsEqual(object x, object y)
    {
        if (x == null && y == null)
        {
            return true;
        }
        if (x == null || y == null)
        {
            return false;
        }
        return XmlUtil.ConvertToXml(x) == XmlUtil.ConvertToXml(y);
    }
}

//the methods from this class are also available at: http://blog.nitriq.com/PutDownTheXmlNodeAndStepAwayFromTheStringBuilder.aspx
public static class XmlUtil
{
    public static string ConvertToXml(object item)
    {
        XmlSerializer xmlser = new XmlSerializer(item.GetType());
        using (System.IO.MemoryStream ms = new System.IO.MemoryStream())
        {
            xmlser.Serialize(ms, item);
            UTF8Encoding textconverter = new UTF8Encoding();
            return textconverter.GetString(ms.ToArray());
        }
    }

    public static T FromXml<T>(string xml)
    {
        XmlSerializer xmlser = new XmlSerializer(typeof(T));
        using (System.IO.StringReader sr = new System.IO.StringReader(xml))
        {
            return (T)xmlser.Deserialize(sr);
        }
    }

}

然后，最后，您可以使用 Fluent.NHibernate 像这样映射您的列：

public partial class MyTableEntityMap: ClassMap<MyTableEntity>
{
    public MyTableEntityMap()
    {
        Table("MyTable");
         //...

        Map(x => x.MyStronglyTypedProperty).Column("SomeXmlTypeSqlColumn").CustomType(typeof(XmlType<TypeOfMyProperty>));
    }
}
于 2012-10-16T18:50:43.103 回答
4

您需要编写一个IUserType负责转换的代码。

您可以从XmlDocType开始，它实际上是从原始 XML 转换为 XmlDocument 的那个。

于 2012-04-04T13:05:21.503 回答