Creating Choice Column in JSGrid Control

Question

Hello, I have created a JSGrid control by using this helpful tutorial:
http://msdn.microsoft.com/en-us/library/ee539027(v=office.14).aspx
However, I was asked to add one "Choice" Column and I haven't figured out a way to do so. Basically what I need is to create a column based on a fixed list of options and be able to select from them by using a dropdown.
Any help will be much appreciated.
Thanks,
Jose M. Cadenas, Msc. Sr. Developer

Answer 1

Found the way to do it...
Hopefully this helps somebody else trying to do the same.
In the formatGridField function... assuming that you have it of course; set the property type for the field you want to make a dropdown.
gf.PropertyTypeId =
"myLookupPropType";
Now, when you create the serializer... assuming that you have something like this:
DataTable data =
new GridData().Data();
// Create a grid serializer to connect to the data.
GridSerializer gds =
new GridSerializer(SerializeMode.Full, data,
"Task ID",
 new
FieldOrderCollection(new
String[] { "Order" }),
 GridUtilities.GetGridFields(data),
GridUtilities.GetGridColumns(data));
Right after creating the GridSerializer, create a list of items as follows.
List<LookupTypeItem> _items =
new List<LookupTypeItem>();
_items.Add(new
LookupTypeItem(1,”Option 1”));
_items.Add(new
LookupTypeItem(2,”Option 2”));
_items.Add(new
LookupTypeItem(3,”Option 3”));
Then create the actual lookup prop type.
LookupTypeInfo lkfield =
new LookupTypeInfo("myLookupPropType", _items);
Now, you just need to register the prop type.
gds.RegisterPropLookupType(lkfield);
and there you go!!!
Jose M. Cadenas, Msc. Sr. Developer