Grouping Functions Within an OCX

[Cenobitez]

Hello Everyone, this is my first post, so i thought i'd say hi first )

Down to my problem now, I am writting an OCX which in the region of 50 Functions and a few subs so far, the final aim to have 100+ least thats how many Idea's i have left approx.....

The Functions will be in sets, the main sections at this point are like,

ControlName.FindWindowName
ControlName.FindLocateBox
ControlName.FindLocateName
ControlName.FindLocateAddress

and then there is the next step like

ControlName.GetWindowName
ControlName.GetLocateBoxText
ControlName.GetLocateName
ControlName.GetLocateAddress

I wanted to be able to group them at first i was thinking by Sections, like Locate, Profile then i determed, that i wanted to group them into like

ControlName.Get.FunctionName
ControlName.Find.FunctionName
ControlName.Set.FunctionName

Much like u can with a text box and

text1.font.bold = true

and

text1.FontBold

Both the above work but u have the option of grouping them like

Text1.Font.Italic
Text1.Font.Bold
Text1.Font.Underline

The order is like ControlName.Group.Function...

I have thought about this, i thought maybe a custom data type, but then they dont really cover this area, Then i thought using an Enumerate, and then i was told collections are the way to do this...

Can this be done in VB or is this a C thing or can it be done in VB ?

Thanks For the help, Im finding out alot just by reading some of the other topics, stuff i hadn't even thought of but i think will be useful...

______
Steve

[peet]

Hi Cenobitez,
add a class module for each group.

add the functions to the classes where they belong.

in u'r usercontrol, add a public varialble for each group, with the name u want to show up when the user hit .

this way u will have code like u wanted.

If u need me to elaborate on this subject, let me know.

[Cenobitez]

Thanks Peet, Let me have a play, if i have no luck then il ask for more, sometimes a shove in the right direction is better than being given it on a plate


_____
Steve

[peet]

Ok