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()](http://stackoverflow.com/questions/784012/javascript-equivalent-of-phps-in-array) — Martijn, Sep 18 '13 at 15:27
Shouldn’t the returned take care of that ? Anyways I tried that, but still no luck :( — Alejandro Reinel, Sep 18 '13 at 15:29
@AlejandroReinel: The `return` *returns* the value. It doesn't let you access private function variables. You want `alert(test())`. — gen_Eric, Sep 18 '13 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, Sep 18 '13 at 15:48

score 4 · Accepted Answer · edited May 23 '17 at 11:56

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 '17 at 11:56

Community

1
1

answered Sep 18 '13 at 15:27

Tushar Gupta - curioustushar

58,085
24
103
107

The return in the `if` will bail out of this function before alerting. – Buggabill Sep 18 '13 at 15:31
I see twice 'var n' ;) – A. Wolff Sep 18 '13 at 15:31
but i am still getting number that are included in my array in the alert – Alejandro Reinel Sep 18 '13 at 15:47
@AlejandroReinel can u explain what you need or make a new question. – Tushar Gupta - curioustushar Sep 18 '13 at 15:53
@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!! – Alejandro Reinel Sep 18 '13 at 15:58

score 1 · Answer 2 · answered Sep 18 '13 at 15:33

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 '13 at 15:33

Sean Burton

907
7
15

okay, i fixed the variable problem, but i thougt the if/else statments where creating a loop – Alejandro Reinel Sep 18 '13 at 15:40
That's not how if/else works. If you want a loop, you need an actual loop, using 'for' or 'do/while' for example. – Sean Burton Sep 20 '13 at 12:44

score 0 · Answer 3 · answered Sep 18 '13 at 15:27

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 '13 at 15:27

tymeJV

103,943
14
161
157

Dosent var make it global? – neoeahit Sep 18 '13 at 15:30
2

No, `var` does not make a global, it simply declares a variable – tymeJV Sep 18 '13 at 15:30

score 0 · Answer 4 · answered Sep 18 '13 at 15:27

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 '13 at 15:27

lukas.pukenis

13,057
12
47
81

score 0 · Answer 5 · answered Sep 18 '13 at 15:29

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
    }

score 0 · Answer 6 · answered Sep 18 '13 at 16:15

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 '13 at 16:15

Akash Sarawagi

207
1
5

Return value if it does not exist in array

6 Answers6