Autoit Random number help?

[TheFixer 1]

So I need some help...

basically all I want to do is have Autoit generate a random number between 1 and 3

and then have an if statement that reads the random number... and lets say if it = 3 it will say the number 3 in a msgbox

(like this)

sleep(500)

Random(1,3)

If Random = 2 Then
    MsgBox(0,"","2")
EndIf

If Random = 1 Then
    MsgBox(0,"","1")
EndIf

If Random = 3 Then
    MsgBox(0,"","3")
EndIf

can anyone help me? (thanks in advance!)

[ahmet 15]

$RandomNumber=Random(1,3,1)
MsgBox(0,"Result",$RandomNumber)

 

[TheFixer 1]

$RandomNumber=Random(1,3,1)
MsgBox(0,"Result",$RandomNumber)

 

not exactly what i was looking for I want it to be like
 

If $RandomNumber = 3 Then
--> does stuff here
EndIf

 

[Melba23 3,790]

TheFixer,

Get the random number and then use a Switch structure to distinguish between the 3 returns - look in the Help file to see the syntax.

M23

[TheFixer 1]

TheFixer,

Get the random number and then use a Switch structure to distinguish between the 3 returns - look in the Help file to see the syntax.

M23

I hope you don't mind but could you give me a little example? (I work best with examples & having things explained to me like I'm stupid pretty much lol)

[Melba23 3,790]

TheFixer,

What do you not understand about the example for Switch in the Help file? Does that not explain how to use the function well enough?

M23

[TheDcoder 1,816]

$iRandomNumber = Random(1, 3, 1)

Switch $iRandomNumber
    Case 1
        ; Code
    
    Case 2
        ; Code
    
    Case 3
        ; Code
EndSwitch

 

Edited September 30, 2015 by TheDcoder
speeling

[TheFixer 1]

$iRandomNumer = Random(1, 3, 1)

Switch $iRandomNumer
    Case 1
        ; Code
    
    Case 2
        ; Code
    
    Case 3
        ; Code
EndSwitch

 

Thank you very much

[TheDcoder 1,816]

My pleasure