对象到 DataView 或 DataSet 或 DataTable 并返回到对象

Question

我们有一个混合应用程序，其遗留模块仍然使用 DataSets、DataViews 和 DataTables，但是我们拥有除此模块的 DB 之外的大多数数据库 ORMed。我想知道是否有人可以指导我如何构建扩展，就像

/* generates a dataset called CustomerDS with 
DataTable called Customer uses property names as DataColumn name */
var dataset =_customer.AsDataSet(); 
/* Converts the dataset to required object or 
throws exception if its cant convert*/
 var customerEntity = _dataset.ToObject<Customer>(); 

我不知道我们什么时候有时间在应用程序的其他层上工作并将其从数据集中释放出来一样。我可能听起来很疯狂，但这只是一个想法。当我需要支持/修复该应用程序的错误时，我会做噩梦。

Answer 1

您可以使用反射，例如：

class Program {
        public static void Start( string[] args ) {
            var john = new Customer {
                CustomerID = Guid.NewGuid(),
                CustomerName = "John",
                CustomerCode = "J-O"
            };

            var tblJohn = john.ToDataTable();
            var clonedJohn = tblJohn.Rows[0].ToDataObject<Customer>();
        }
}

[AttributeUsage(AttributeTargets.Property)]
public class DataColumnAttribute : Attribute { }
public class Customer {
    [DataColumn]
    public Guid CustomerID { get; set; }

    [DataColumn]
    public string CustomerName { get; set; }

    [DataColumn]
    public string CustomerCode { get; set; }
}

public static class DataObjectExtensions {
    public static T ToDataObject<T>( this DataRow dataRow ) where T : new() {
        var dataObject = Activator.CreateInstance<T>();
        var tpDataObject = dataObject.GetType();

        foreach ( var property in tpDataObject.GetProperties() ) {
            var attributes = property.GetCustomAttributes( typeof( DataColumnAttribute ), true );
            if ( null != attributes && attributes.Length > 0 ) {
                if ( property.CanWrite ) {
                    DataColumn clm = dataRow.Table.Columns[property.Name];
                    if ( null != clm ) {
                        object value = dataRow[clm];
                        property.SetValue( dataObject, value, null );
                    }
                }
            }
        }

        return dataObject;
    }
    public static DataTable ToDataTable( this object dataObject ) {
        var tpDataObject = dataObject.GetType();

        DataTable tbl = new DataTable();
        DataRow dataRow = tbl.NewRow();
        foreach ( var property in tpDataObject.GetProperties() ) {
            var attributes = property.GetCustomAttributes( typeof( DataColumnAttribute ), true );
            if ( null != attributes && attributes.Length> 0 ) {
                if ( property.CanRead ) {
                    object value = property.GetValue( dataObject, null );
                    DataColumn clm = tbl.Columns.Add( property.Name, property.PropertyType );
                    dataRow[clm] = value;
                }
            }
        }

        tbl.Rows.Add( dataRow );
        tbl.AcceptChanges();
        return tbl;
    }
}

Answer 2

您可以使用它从数据表中获取对象

 public static class Extensions
    {
        public static List<T> ToList<T>(this DataTable table) where T : new()
        {
            IList<PropertyInfo> properties = typeof(T).GetProperties().ToList();
            List<T> result = new List<T>();

            foreach (var row in table.Rows)
            {
                var item = CreateItemFromRow<T>((DataRow)row, properties);
                result.Add(item);
            }

            return result;
        }

        private static T CreateItemFromRow<T>(DataRow row, IList<PropertyInfo> properties) where T : new()
        {
            T item = new T();
            foreach (var property in properties)
            {
                if (property.PropertyType == typeof(System.DayOfWeek))
                {
                    DayOfWeek day = (DayOfWeek)Enum.Parse(typeof(DayOfWeek), row[property.Name].ToString());
                    property.SetValue(item,day,null);
                }
                else
                {
                    property.SetValue(item, row[property.Name], null);
                }
            }
            return item;
        }
    }

并像这样使用它

List<Employee> lst = ds.Tables[0].ToList<Employee>();