CODEWITHSUNDEEP
javascripthtmlrandom
AisRuss
AisRuss

Reputation: 269

Javascript random number not displaying result

I've got a webpage which is meant to generate a random number, then when the number =5 it displays a win message..if not display lose message, but its not displaying any alerts..have i missed something out?

<!DOCTYPE html>
<html>
<head>
<script type="text/javascript">

function WinLose()
{
var x=document.getElementById("demo");
x=x.innerHTML=Math.floor((Math.random()*5)+1);
return x;
if (x=4)
{
    alert("winner!");
}
else
{
    alert("loser");
}
}
</script>

</head>
<body>


<p id="demo">Click the button to display a random number between 1 and5.</p>


<button onclick="WinLose()">Try it</button>

</body>
</html>

EDIT: Managed to get this bit working so now it displays either win or loose depending on its number, yet does anyone know how i can swap the alerts in the if statements to display a DIV section. ive got a jQuery file included so it can accept the hide/show effect...anything i tried didnt work

Upvotes: 0

Views: 630

Answers (5)

technosaurus
technosaurus

Reputation: 7802

Here is a simplified version that externalizes the alerts (I assume you don't plan to use them later)

function WinLose(){
  var x=Math.floor((Math.random()*5)+1);
  document.getElementById("demo").innerHTML=x;
  return (x==5) ? "Winner!" : "Loser";
}

alert(WinLose());

Upvotes: 0

Luiz Edgar
Luiz Edgar

Reputation: 71

Your code:

x=x.innerHTML=Math.floor((Math.random()*5)+1);

x is receiving x.innerHTML then x.innerHTML = random number.

Correct way: (remove "x=")

x.innerHTML = Math.floor((Math.random()*5)+1);

Upvotes: 0

Eric Goncalves
Eric Goncalves

Reputation: 5353

You need to return x after you generate the random value; and you must add x == 4 

function WinLose()
{
var x=document.getElementById("demo");
x=x.innerHTML=Math.floor((Math.random()*5)+1);
if (x==4) {
    alert("winner!");
} else {
    alert("loser");
}
    return x;
}

DEMO

Upvotes: 0

Kyle Gagnon
Kyle Gagnon

Reputation: 62

Yeah, It can be tough. The main problem again has to be other than return x is the "==". So this

if (x=4)

Should really say:

if (x==4)

What you said before was that you were assinging x to 4 so that has no meaning at all and messes everything up.

Hope this helps you!

Upvotes: 2

Mike Corcoran
Mike Corcoran

Reputation: 14565

you have return x after you generate a random value for x. this means no javascript code after that line will run in that function.

also, your if statement needs to use '==' to do the comparison rather than '=' which is assignment.

Upvotes: 3

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