dapper.fastcrud并未映射来自postgresql的几何数据

时间:2018-10-06 18:27:41

标签: postgresql dapper spatial dapper-fastcrud

我在Postgresql中有一个空间数据。例如,表格planet_osm_point具有2个属性:

CREATE TABLE public.planet_osm_point
(
    osm_id bigint,
    way geometry(Point,3857)
)

如果我使用dapper进行CRUD操作,则一切正常。但是,如果我使用Dapper.fastCRUD,则带有几何图形的“ way”属性始终为空

OsmPoint类:

using NetTopologySuite.Geometries;
using System.ComponentModel.DataAnnotations.Schema; 

namespace DapperTest
{
    [Table("planet_osm_point")]
    public class OsmPoint
    {
        [Column("osm_id")]
        public long OsmId { get; set; }

        [Column("way")]
        public Point Way { get; set; }
    }
}

如果我使用Dapper,那么我会收到Way属性具有几何坐标:

using (NpgsqlConnection conn = new NpgsqlConnection(_connectionString))
{
    conn.Open();
    conn.TypeMapper.UseNetTopologySuite();
    var result = conn.Query<OsmPoint>("SELECT * FROM planet_osm_point LIMIT 5000").ToList();
    return result;
}

但是如果我使用Dapper.fastCRUD,则Way始终为空

using (NpgsqlConnection conn = new NpgsqlConnection(_connectionString))
{
    conn.Open();
    conn.TypeMapper.UseNetTopologySuite();
    var result = conn.Find<OsmPoint>(p=>p.Top(5000));
    return result;
}

有人知道如何使Dapper.fastCRUD处理几何数据吗?

2 个答案:

答案 0 :(得分:0)

首先,您要为Geometry类型创建TypeHandler,如下所示:

public class GeometryTypeMapper : SqlMapper.TypeHandler<Geometry>
{
    public override void SetValue(IDbDataParameter parameter, Geometry value)
    {
        if (parameter is NpgsqlParameter npgsqlParameter)
        {
            npgsqlParameter.NpgsqlDbType = NpgsqlDbType.Geometry;
            npgsqlParameter.NpgsqlValue = value;
        }
        else
        {
            throw new ArgumentException();
        }
    }

    public override Geometry Parse(object value)
    {
        if (value is Geometry geometry)
        {
            return geometry;
        }

        throw new ArgumentException();
    }
}

添加类型处理程序:

SqlMapper.AddTypeHandler(new GeometryTypeMapper());

在FastCURD查询之前设置属性:

OrmConfiguration.GetDefaultEntityMapping<OsmPoint>().SetProperty(p => p.way, prop => prop.SetDatabaseColumnName("way"));

答案 1 :(得分:0)

Dapper.FastCRUD作为默认值仅使用对称SQL类型来构建查询。 您可以在Dapper.FastCrud.Configuration.OrmConventions.cs中看到它:

public virtual IEnumerable<PropertyDescriptor> GetEntityProperties(Type entityType)
{
    return TypeDescriptor.GetProperties(entityType)
        .OfType<PropertyDescriptor>()
        .Where(propDesc => 
            !propDesc.Attributes.OfType<NotMappedAttribute>().Any()
            && !propDesc.IsReadOnly 
            && propDesc.Attributes.OfType<EditableAttribute>().All(editableAttr => editableAttr.AllowEdit)
            && this.IsSimpleSqlType(propDesc.PropertyType));
}

我重写IsSimpleSqlType(Type propertyType)方法:

public class GeometryConvention: OrmConventions
{
    protected override bool IsSimpleSqlType(Type propertyType)
    {
        var res = base.IsSimpleSqlType(propertyType) || propertyType.BaseType != null && propertyType.BaseType.Name.StartsWith("geometry", StringComparison.OrdinalIgnoreCase);
        return res;
    }
}

注册自定义约定:

OrmConfiguration.Conventions = new GeometryConvention();

并使用自定义的GeometryTypeMapper,例如在LongNgôThành答案中。