Calling a variable within another function

[KyleB]

I'm creating a script to create a new user and log into ts every time, I had it working but when i started simplifying my code into functions so i could separate them while testing, It couldn't find the variable. I'm just wondering if there is a way to take that variable from the other function.

 

LaunchSEQ ()
NameGen ()
JoinSEQ ()

Func NameGen ()
Local $arr[26] = ["Alpha","Bravo","Charlie","Delta","Echo","Foxtrot","Golf","Hotel","India","Juliet","Kilo","Lima","Mike","November","Oscar","Papa","Quebec","Romeo","Sierra","Tango","Uniform","Victor","Whisky","X-Ray","Yankee","Zulu"]
   $i = Random(1, 26, 1)
   $name = $arr[$i] & "|Burns"
Send($name)
EndFunc

Func JoinSEQ ()
send("{ENTER}")
Send("^s")
Sleep(250)
Send("{TAB}")
Sleep(250)
Send("^a")
Sleep(250)
Send($name)
sleep(250)
send("{ENTER}")
EndFunc

Error is located at the bottom regarding the variable "$name"

[JLogan3o13]

I don't see the error you reference, but if you need $name to be available to the entire script, the declare it at the top as Global. If this function is going to return that variable for use in a subsequent function, use Return instead (see help file for both).

Edited November 11, 2016 by JLogan3o13

[KyleB]

 

It says "Error: Variable used without being declared.", while i have declared it above it, just within another function.

   $name = $arr[$i] & "|Burns"
Send($name)
EndFunc

Sleep(250)
Send($name)
sleep(250)
send("{ENTER}")
EndFunc

Edited November 11, 2016 by KyleB

[JLogan3o13]

Yes, that is the code - I thought you were posting the actual error you're receiving. That will help us a great deal.

[KyleB]

Fixed it, thanks

[spudw2k]

On 11/11/2016 at 1:46 PM, KyleB said:

Fixed it, thanks

It's always helpful to the forum community if you share how you resolved it.   Was it the because your random function upper and lower limits were wrong (zero-indexed array...0 to 25 instead of 1 to 26)?