Reputation: 14692
I'm having a Structure like
X={ID="1", Name="XX",
ID="2", Name="YY" };
How to dump this data to a DataGridView
of two columns
The gridView is like
ID | Name
Can we use LINQ to do this. I'm new to DataGridView
Pleaese help me to do this..
Thanks in advance
Upvotes: 12
Views: 134938
Reputation: 5480
Let's assume you have a class like this:
public class Staff
{
public int ID { get; set; }
public string Name { get; set; }
}
And assume you have dragged and dropped a DataGridView
to your form, and name it dataGridView1.
You need a BindingSource to hold your data to bind your DataGridView
. This is how you can do it:
private void frmDGV_Load(object sender, EventArgs e)
{
//dummy data
List<Staff> lstStaff = new List<Staff>();
lstStaff.Add(new Staff()
{
ID = 1,
Name = "XX"
});
lstStaff.Add(new Staff()
{
ID = 2,
Name = "YY"
});
//use binding source to hold dummy data
BindingSource binding = new BindingSource();
binding.DataSource = lstStaff;
//bind datagridview to binding source
dataGridView1.DataSource = binding;
}
Upvotes: 5
Reputation: 2752
My favorite way to do this is with an extension function called 'Map':
public static void Map<T>(this IEnumerable<T> source, Action<T> func)
{
foreach (T i in source)
func(i);
}
Then you can add all the rows like so:
X.Map(item => this.dataGridView1.Rows.Add(item.ID, item.Name));
Upvotes: 3
Reputation: 6184
you shoud do like this form your code
DataGridView.DataSource = yourlist;
DataGridView.DataBind();
Upvotes: 0
Reputation: 5291
LINQ is a "query" language (thats the Q), so modifying data is outside its scope.
That said, your DataGridView
is presumably bound to an ItemsSource
, perhaps of type ObservableCollection<T>
or similar. In that case, just do something like X.ToList().ForEach(yourGridSource.Add)
(this might have to be adapted based on the type of source in your grid).
Upvotes: 0
Reputation: 7361
first you need to add 2 columns to datagrid. you may do it at design time. see Columns property. then add rows as much as you need.
this.dataGridView1.Rows.Add("1", "XX");
Upvotes: 30