Launch different programs based on time.

[jubjub]

The below isn't working as expected. At 5:11 PM it is launching unilaunchFirstShift.exe and unilaunchThirdShift.exe.

 

I put the message box in to debug and it verifies that @Hour is 17, so why does this script think 17 is <= 5 and >= 0?

Spoiler

#include <Date.au3>
 

Local $Time=@HOUR

MsgBox(0, "", $Time )

     If @Hour <= "5" And @Hour >= "0" Then
Run ("C:\util\unilaunchThirdShift.exe")

   EndIf

     If @Hour <= "11" And @Hour >= "6" Then
Run ("C:\util\unilaunchFourthShift.exe")

   EndIf

If @Hour <= "17" And @Hour >= "12" Then
Run ("C:\util\unilaunchFirstShift.exe")

   EndIf

   If @Hour <= "23" And @Hour >= "18" Then
Run ("C:\util\unilaunchSecondShift.exe")

   EndIf

 

 

Edited February 27, 2019 by jubjub

[jubjub]

Update: When the hour turned to 6:00 (18) it is now launching second shift and third shift. Why is third shift always matching?

 

Update to the update:

If I change it to If...ElseIf...Elseif...Else and put third shift last it appears to be working properly, but that doesn't tell my why my logic wasn't working without having to "hack" it.

Edited February 27, 2019 by jubjub

[Nine]

Because you are mixing numbers with strings.  Read carefully help file : @HOUR vary from "00" to "23".  "0", "5","6" are invalid in that context.  Now your homework is to find the right value for those three.

[jubjub]

32 minutes ago, Nine said:

Because you are mixing numbers with strings.  Read carefully help file : @HOUR vary from "00" to "23".  "0", "5","6" are invalid in that context.  Now your homework is to find the right value for those three.

Thank you very much. 

[jchd]

Try this:

Local $aShift = ["Third", "Fourth", "First", "Second"]
Run("C:\util\unilaunch" & $aShift[Mod(@HOUR, 6)] & "Shift.exe")