I have a method that I call to create a link button. However the problem is that it passing the field name that the actual field to link button becasically giving me a dead link.
This is the method:
private static DataGridTemplateColumn CreateHyperlink(string fieldName)
{
DataGridTemplateColumn column = new DataGridTemplateColumn();
column.Header = "";
string link = @"http://www.amazon.com/gp/product/" + fieldName;
StringBuilder sb = new StringBuilder();
sb.Append("<DataTemplate ");
sb.Append("xmlns='http://schemas.microsoft.com/winfx/2006/xaml/presentation' ");
sb.Append("xmlns:x='http://schemas.microsoft.com/winfx/2006/xaml' ");
sb.Append("xmlns:src='clr-namespace:Silverlight1.Classes;assembly=Silverlight1.Classes'>");
sb.Append("<HyperlinkButton ");
sb.Append("TargetName= '_blank' ");
sb.Append("Content = 'Link' ");
sb.Append("NavigateUri =" +"'"+ link +"'");
sb.Append(" />");
sb.Append("</DataTemplate>");
column.CellTemplate = (DataTemplate)XamlReader.Load(sb.ToString());
column.IsReadOnly = false;
return column;
}
Which is called by this
dgOrder.Columns.Add(CreateHyperlink("asin"));
Which is getting pulled from a WCF Silverlight enable data service.
How do I pass the content instead of field name?
You don’t pass the content, you pass the field name but create a
Binding. You will also need an implementation of aIValueConverter.Lets start with an
IValueConverter, you need something which take the string value of the “field” (by which I’m assuming you mean a Property of the objects bound to rows in the grid) and prefix it with@"http://www.amazon.com/gp/product/"to form a complete URL.Place an instance of this in the App.xaml:-
Now your create method can look like this:-