Reputation: 17298
Hello again; i need to show X,Y,Risk in ListBoxes. But i can not do it.
public partial class _Default : System.Web.UI.Page
{
protected void Page_Load(object sender, EventArgs e)
{
if (!IsPostBack)
{
List<TrainigSet> listtraining = new List<TrainigSet>();
TrainigSet tr = new TrainigSet();
double[] X = { 2, 3, 3, 4, 5, 6, 7, 9, 11, 10 };
double[] Y = { 4, 6, 4, 10, 8, 3, 9, 7, 7, 2 };
string[] Risk = { "Kötü", "iyi", "iyi", "kötü", "kötü", "iyi", "iyi", "kötü", "kötü", "kötü" };
for (int i = 0; i < X.Length; i++)
{
tr.X = X[i];
tr.Y = Y[i];
tr.Risk = Risk[i];
listtraining.Add(tr);
}
for (int i = 0; i < listtraining.Count; i++)
{
ListBox1.Items.Add(listtraining[i].X.ToString());
ListBox2.Items.Add(listtraining[i].Y.ToString());
ListBox3.Items.Add(listtraining[i].Risk.ToString());
}
}
}
}
public class TrainigSet
{
public double X { get; set; }
public double Y { get; set; }
public string Risk { get; set; }
}
Upvotes: 1
Views: 6258
Reputation: 2987
You could also use a TrainingResult class with a public X,Y, and Risk like this
public class TrainingResult
{
public double X{get;set;}
public double Y{get;set;}
public string Risk {get;set};
}
And create a list of those. Next you could bind to it like this:
ListBoxX.DataSource = List<TrainingResult>
ListBoxX.DataMember = "X";
ListBoxX.DataBind();
ListBoxX.DataSource = List<TrainingResult>
ListBoxY.DataMember = "Y";
ListBoxY.DataBind();
ListBoxRisk.DataSource = List<TrainingResult>
ListBoxRisk.DataMember = "Risk";
ListBoxRisk.DataBind();
The advantage of this being that you have a more clear relation between x,y and risk and get more readable (to me) code. Disadvantage is offcourse the DataMember being a string value.
Upvotes: 1
Reputation: 96596
You have to move the instantiation/creation of the TrainingSet into the for loop (you want to create a new instance for every item you add to listtraining):
double[] X = { ... };
double[] Y = { ... };
string[] Risk = { ... };
for (int i = 0; i > X.Length; i++)
{
TrainigSet tr = new TrainigSet(); // create a new TrainingSet
...
listtraining.Add(tr);
}
Otherwise you will modify the same TrainingSet instance over and over again.
Upvotes: 1