You can't query against the DataTable
's Rows collection, since DataRowCollection
doesn't implement IEnumerable<T>
. You need to use the AsEnumerable()
extension for DataTable
. Like so:
var results = from myRow in myDataTable.AsEnumerable()
where myRow.Field<int>("RowNo") == 1
select myRow;
And as @Keith says, you'll need to add a reference to System.Data.DataSetExtensions
AsEnumerable()
returns IEnumerable<DataRow>
. If you need to convert IEnumerable<DataRow>
to a DataTable
, use the CopyToDataTable()
extension.
Below is query with Lambda Expression,
var result = myDataTable
.AsEnumerable()
.Where(myRow => myRow.Field<int>("RowNo") == 1);
Whereas one approach is to implement the ICloneable
interface (described here, so I won't regurgitate), here's a nice deep clone object copier I found on The Code Project a while ago and incorporated it into our code.
As mentioned elsewhere, it requires your objects to be serializable.
using System;
using System.IO;
using System.Runtime.Serialization;
using System.Runtime.Serialization.Formatters.Binary;
/// <summary>
/// Reference Article http://www.codeproject.com/KB/tips/SerializedObjectCloner.aspx
/// Provides a method for performing a deep copy of an object.
/// Binary Serialization is used to perform the copy.
/// </summary>
public static class ObjectCopier
{
/// <summary>
/// Perform a deep copy of the object via serialization.
/// </summary>
/// <typeparam name="T">The type of object being copied.</typeparam>
/// <param name="source">The object instance to copy.</param>
/// <returns>A deep copy of the object.</returns>
public static T Clone<T>(T source)
{
if (!typeof(T).IsSerializable)
{
throw new ArgumentException("The type must be serializable.", nameof(source));
}
// Don't serialize a null object, simply return the default for that object
if (ReferenceEquals(source, null)) return default;
using var Stream stream = new MemoryStream();
IFormatter formatter = new BinaryFormatter();
formatter.Serialize(stream, source);
stream.Seek(0, SeekOrigin.Begin);
return (T)formatter.Deserialize(stream);
}
}
The idea is that it serializes your object and then deserializes it into a fresh object. The benefit is that you don't have to concern yourself about cloning everything when an object gets too complex.
In case of you prefer to use the new extension methods of C# 3.0, change the method to have the following signature:
public static T Clone<T>(this T source)
{
// ...
}
Now the method call simply becomes objectBeingCloned.Clone();
.
EDIT (January 10 2015) Thought I'd revisit this, to mention I recently started using (Newtonsoft) Json to do this, it should be lighter, and avoids the overhead of [Serializable] tags. (NB @atconway has pointed out in the comments that private members are not cloned using the JSON method)
/// <summary>
/// Perform a deep Copy of the object, using Json as a serialization method. NOTE: Private members are not cloned using this method.
/// </summary>
/// <typeparam name="T">The type of object being copied.</typeparam>
/// <param name="source">The object instance to copy.</param>
/// <returns>The copied object.</returns>
public static T CloneJson<T>(this T source)
{
// Don't serialize a null object, simply return the default for that object
if (ReferenceEquals(source, null)) return default;
// initialize inner objects individually
// for example in default constructor some list property initialized with some values,
// but in 'source' these items are cleaned -
// without ObjectCreationHandling.Replace default constructor values will be added to result
var deserializeSettings = new JsonSerializerSettings {ObjectCreationHandling = ObjectCreationHandling.Replace};
return JsonConvert.DeserializeObject<T>(JsonConvert.SerializeObject(source), deserializeSettings);
}
Best Answer
There's no way to create a table automatically from the
DataGridView
columns (EDIT To refine this: There are ways to automatically create database structures - ORM would be a good keyword here - but I guess that's not what you need). The other way round works well: Create the table and assign it to theDataGridView
as its data source. Then the grid should show the respective columns."Storing" the values in the data table works only if you have "told" the columns, which field of the underlying table they relate to. That is: You need to create the table columns (it has a name), and when you create the
DataGridView
column, you need to assign the column name to theDataPropertyName
property of the column.Easiest way for you: Create a typed dataset (use the "Add to project" dialog to add a new "DataSet"), manually create a table in the dataset's designer, drag an instance of your dataset to the form and assign this as the data source to your
DataGridView
. Everything else should appear automatically.