Hi All
I am trying to configure a Job Registration terminal to be used as a Clocking In and Out terminal.
The process needs to be as quick as possible as we dont want queues forming.
It will not be used for job registration etc - purely clocking in and out.
I can make a clocking in terminal easily, but it displays a popup box which has to be Ok’d (or disappears after 10 seconds).
First question - is there any way to get this popup box to disappear after e.g.1-2 seconds instead?
Second question - If a user clocks in it informs them that they are clocked in. If they clock in it says the same message - It should inform them on the second time that they are already clocked in.
For the clock out it need to be as quick as possible, but ideally warn them or not allow them to clock out unless they have reported a qty on the job they were on.
Has anyone done anything similar?
Thanks in advance.
You can execute a method (which may close a form, for example) with the help of setTimeOut().
For example, if you call the following code in a form, it will close the form after two seconds:
this.setTimeOut(methodStr(FormRun, close), 2000);
Hi - I really am new to this.
How do I do that?
What exactly do you want to know?
I would like to know how to add the custom code you suggested to make the ‘You have not clocked in’ popup to close in a faster time than 10 seconds.
I am trying to make a clocking terminal using the standard screens where the user can literally clock in and clock out as quickly as possible without having lots of buttons etc to press.
Just copy the code to init() method (for example) of any form.
Hi
How do I actually get to the code … I am very new to this.
I know how to log in and get to AOT 
Aha, you should have started on the beginning…
What’s your version of AX? Do you have a development license and permissions?
By the way, you’ll find many answers on MSDN. You can’t learn everything needed just by asking on forums.
Hi
We are AX 2012 R2.
I think we have the development licence and I have full permissions.