SkellySoul Posted September 9, 2009 Share Posted September 9, 2009 (edited) I have already did a search and found slot machines in the example scripts but they are more complicated than I want them to be. This is the code I got so far and I was just wondering if this works how I want it to more or less I just need a second eye to help spot any errors. #include <GUIConstantsEx.au3> #include <WindowsConstants.au3> Global $Wheel[4] $Slots = GUICreate("Form1", 600, 600) $Wheel[1] = GUICtrlCreatePic("1.jpg", 0 , 0 , 200 , 200) $Wheel[2] = GUICtrlCreatePic("1.jpg", 200 , 0 , 200 , 200) $Wheel[3] = GUICtrlCreatePic("1.jpg", 400 , 0 , 200 , 200) $Spin = GUICtrlCreateButton("Spin" , 0 , 200 , 600 , 20) GUISetState(@SW_SHOW) While 1 $Msg = GUIGetMsg() Switch $Msg Case $GUI_EVENT_CLOSE Exit Case $Spin _Spin() EndSwitch WEnd Func _Spin() $Delay = 0 Sleep($Delay) $iWheel = 1 Do $Random = Random(1 , 4 , 1) If $iWheel = 2 And $Random = 4 Then $iRandom = 5 Else $iRandom = $Random EndIf GUICtrlSetImage($Wheel[$iWheel] , $iRandom & ".jpg") $iWheel += 1 Until $iWheel = 4 EndFunc The reason why if it is on wheel 2 and random 4 it returns a 5 because 5 is a wild and a wild can only be found in the center of the slot machine. This code seems to work for me. If this helps explain it any better Min Bid = 1 Max Bid = 3 Payout = 2 * Bid = 2,4,6 Pay out 1 Bid = 2 Pay out 2 Bid = 4 Pay out 3 Bid = 6 Final Payout Table: (1) (1) (1) = 2 * Bid (2) (2) (2) = 3 * Bid (3) (3) (3) = 4 * Bid (4) (5) (4) = 5 * Bid (1) (5) (1) = 2 * Bid (2) (5) (2) = 3 * Bid (3) (5) (3) = 4 * Bid (4) (5) (4) = 5 * Bid Edited September 9, 2009 by SkellySoul Link to comment Share on other sites More sharing options...
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