Question

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，但我希望在映射时将XML序列化/反序列化为StronglyTypedData属性。

我需要做些什么来完成这项工作？

Answer 1

我打算对迭戈的帖子做一个评论，但它太长了，我想要语法高亮。我修改了Diego发布的XmlDocType，以便它可以在强类型对象中使用xml序列化。

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映射列，如下所示：

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

        Map(x => x.MyStronglyTypedProperty).Column("SomeXmlTypeSqlColumn").CustomType(typeof(XmlType<TypeOfMyProperty>));
    }
}

Answer 2

您需要编写负责转换的IUserType。

您可以从XmlDocType开始，这是实际从原始XML转换为XmlDocument的那个。

如何使用NHibernate将XML类型列映射到强类型对象属性？

2 个答案: