Return value if it does not exist in array

Question

I need this code to return a different value (from 0 to 14) if its not defined in my array. I dont understan why it tells me that "n" is not defined. please help

EDIT ***

I fixed the n value, but I’m still getting numbers that are included in my array !!

EDIT 2 *******

I need my code to return a numerical value (at random) that is not included in my array. This is a simplified version of what I’m doing, the full version has non sequential values and a lot of numbers!!

function test () {
   var arr= [1,2,3,4,5,6,7,8,9];
   var n = Math.floor((Math.random() * 15));
   var tex = $.inArray( n, arr );
   if (tex == -1) {
       return n;
   }
   else {
       var n = Math.floor((Math.random() * 15));
   }
  alert (n);
}

possible duplicate of JavaScript equivalent of PHP's in_array() — Martijn
– Martijn, Commented Sep 18, 2013 at 15:27
Shouldn’t the returned take care of that ? Anyways I tried that, but still no luck :( — Alejandro Reinel
– Alejandro Reinel, Commented Sep 18, 2013 at 15:29
@AlejandroReinel: The return returns the value. It doesn't let you access private function variables. You want alert(test()). — gen_Eric
– gen_Eric, Commented Sep 18, 2013 at 15:35
@Martijn not a duplicate, because I need a number that is not included in my array, not only to check if its there — Alejandro Reinel
– Alejandro Reinel, Commented Sep 18, 2013 at 15:48

Community · Accepted Answer · 2017-05-23 11:56:35Z

4

function test () {
   var arr= [1,2,3,4,5,6,7,8,9];
   var n = Math.floor((Math.random() * 15));
   var tex = $.inArray( n, arr );
   if (tex == -1) {
       return n;
   }
   else {
       n = Math.floor((Math.random() * 15)); //no need of declaring n again
   }
   alert (n);//alert should be here inside the function 
}

n is defined inside the function test

so var n scope is only inside the function i.e n is not accessible outside the function test

in your code alert (n); will only work if the function fails if condition.As if conditions becomes true it will returns n control will be out of the function i.e no further execution of the code inside the function.

Read --> What is the scope of variables in JavaScript?

Updated after OP's comment

DEMO

function test() {
    var arr = [1, 2, 3, 4, 5, 6, 7, 8, 9];
    var n = Math.floor((Math.random() * 15));
    var tex;
    while ((tex = $.inArray(n, arr)) != -1) {
        n = Math.floor((Math.random() * 15));
    }
    return n;
}
alert(test());

edited May 23, 2017 at 11:56

CommunityBot

11 silver badge

answered Sep 18, 2013 at 15:27

Tushar Gupta - curioustushar

57.2k24 gold badges106 silver badges110 bronze badges

Sign up to request clarification or add additional context in comments.

5 Comments

Buggabill Over a year ago

The return in the if will bail out of this function before alerting.

A. Wolff Over a year ago

I see twice 'var n' ;)

Alejandro Reinel Over a year ago

but i am still getting number that are included in my array in the alert

Tushar Gupta - curioustushar Over a year ago

@AlejandroReinel can u explain what you need or make a new question.

Alejandro Reinel Over a year ago

@TusharGupta I need my code to return a numerical value (at random) that is not included in my array. This is a simplified version of what I’m doing, the full version has non sequential values and a lot of numbers!!

Sean Burton · Accepted Answer · 2013-09-18 15:33:04Z

1

Aside from the fact that the variable goes out of scope when the function ends, as everyone else has mentioned, your function does not do what you say it is supposed to do. If the first number it generates is in the array it only makes one more attempt at generating a random number, then uses this number regardless of whether it is in the array or not. You need some sort of loop to keep attempting new random numbers until it finds one that is definitely not in the array.

answered Sep 18, 2013 at 15:33

Sean Burton

9697 silver badges15 bronze badges

2 Comments

Alejandro Reinel Over a year ago

okay, i fixed the variable problem, but i thougt the if/else statments where creating a loop

Sean Burton Over a year ago

That's not how if/else works. If you want a loop, you need an actual loop, using 'for' or 'do/while' for example.

tymeJV · Accepted Answer · 2013-09-18 15:27:12Z

0

n is local to the function it is defined in, define n outside your function to use it outside the function.

answered Sep 18, 2013 at 15:27

tymeJV

105k14 gold badges165 silver badges158 bronze badges

2 Comments

neoeahit Over a year ago

Dosent var make it global?

tymeJV Over a year ago

No, var does not make a global, it simply declares a variable

lukas.pukenis · Accepted Answer · 2013-09-18 15:27:52Z

0

Your n variable is defined inside a function scope thus is not accesible from the global scope.

To return values from function use this syntax:

return n;

so then you can do something like this in global scope:

alert(function(args));

answered Sep 18, 2013 at 15:27

lukas.pukenis

13.7k13 gold badges50 silver badges83 bronze badges

Comments

neoeahit · Accepted Answer · 2013-09-18 15:29:41Z

0

Firstly you mentioned that you want to return a value: why are you making it global?

Try this:

 function test () {
       var arr= [1,2,3,4,5,6,7,8,9];
       n = Math.floor((Math.random() * 15));
       var tex = $.inArray( n, arr );
       if (tex == -1) {
           return n;
       }
       else {
          n = Math.floor((Math.random() * 15));
       }
return n
    }

answered Sep 18, 2013 at 15:29

neoeahit

1,7992 gold badges22 silver badges36 bronze badges

Comments

Akash Sarawagi · Accepted Answer · 2013-09-18 16:15:08Z

0

function test() {
    var arr = [1, 2, 3, 4, 5, 6, 7, 8, 9];
    var n = Math.floor((Math.random() * 15));
    var tex = $.inArray(n, arr);
    if (tex == -1) {
        return n;
    } else {
        test();
    }
}

answered Sep 18, 2013 at 16:15

Akash Sarawagi

2071 silver badge5 bronze badges

Collectives™ on Stack Overflow

Return value if it does not exist in array

6 Answers 6

5 Comments

2 Comments

2 Comments

Comments

Comments

Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

6 Answers 6

5 Comments

2 Comments

2 Comments

Comments

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related