InfoPath Dev

Hello all, I am new to this forum and new to infopath
I am trying to generate a random number, that will be place into the ID field, everytime I load a form. So what I mean is when I load the form the random number will show into the ID field and the user finally I want the user to not be able to modify this ID field.

ANy help pleaseeee I really aprecciatte

Thank you

Hello
Welcome to InfopathDev Forums

You can do this by generating a GUID which will be a unique number
Try reading the links below
http://www.infopathdev.com/forums/topic.asp?TOPIC_ID=2431
http://www.infopathdev.com/forums/topic.asp?TOPIC_ID=448

Also you can search for more in the forums by typing GUID in the search box above

Maybe, u can use this code:

var value = Math.random()*1000;

regards,

Hello, thank you very much for your help. But where do I put this code? Once I had all my data field in the form I tried to go to Tools>Programming but I cant click in any of those options. Where do I need to write that code, and is just a copy-paste or do I need to write something else?

Ok, now I can see the programming option. I clicked on On Load Event, then I tried putting different lines of code. The field on my ID is called "ID" (I created a db in Access and add a field for each table name: ID). I then drag and drop all the fields from one of my table to the form, so the ID field is in the form. My question is, how do I write this code, where do I put the "ID" name in the code. Is it like: var ID =
new ActiveXObject("Scriptlet.TypeLib").guid.substring( 0, 38); or var ID = System.Guid.NewGuid(); ? is that all the code I need to write? I am so confused, sorry for been such a novice. Thank youu

try this:

Put a button control, go to properties and edit code, put the next code(jscript):


XDocument.DOM.selectSingleNode("/myFields/ID").text = Math.random()*1000;