Using OR operator in a jQuery if statement

Question

I need to use the OR operator in a jQuery if statement to filter out 10 states. My code works wile only excluding one state, but fails when I try to include multiple states. Is there a correct way to do this?

Code I am using :

if ((state != 10) || (state != 15) || (state != 19) || 
    (state != 22) || (state != 33) || (state != 39) || 
    (state != 47) || (state != 48) || (state != 49) || 
    (state != 51)) 
   return true;

@aw04 Yes! using the AND operator worked. I should have tried this before i posted this question. The other answers were good to know, but this was the easiest possible solution. Thanks — Ledattack
– Ledattack, Commented Oct 24, 2016 at 20:24
Does this answer your question? Why does non-equality check of one variable against many values always return true? — khelwood
– khelwood, Commented Jan 14, 2022 at 20:02

EJoshuaS - Stand with Ukraine · Accepted Answer · 2021-07-18 18:52:51Z

33

Think about what

if ((state != 10) || (state != 15) || (state != 19) || (state != 22) || (state != 33) || (state != 39) || (state != 47) || (state != 48) || (state != 49) || (state != 51))

means. || means "or." The negation of this is (by DeMorgan's Laws):

state == 10 && state == 15 && state == 19...

In other words, the only way that this could be false is if a state equals 10, 15, and 19 (and the rest of the numbers in your or statement) at the same time, which is impossible.

Thus, this statement will always be true. State 15 will never equal state 10, for example, so it's always true that state will either not equal 10 or not equal 15.

Change || to &&.

Also, in most languages, the following:

if (x) {
  return true;
}
else {
  return false;
}

is not necessary. In this case, the method returns true exactly when x is true and false exactly when x is false. You can just do:

return x;

edited Jul 18, 2021 at 18:52

answered Oct 24, 2016 at 16:17

EJoshuaS - Stand with Ukraine

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Comments

Venkata Krishna · Accepted Answer · 2016-10-24 16:56:16Z

5

The code you wrote will always return true because state cannot be both 10 and 15 for the statement to be false. if ((state != 10) && (state != 15).... AND is what you need not OR.

Use $.inArray instead. This returns the index of the element in the array.

JSFIDDLE DEMO

var statesArray = [10, 15, 19]; // list out all

var index = $.inArray(state, statesArray);

if(index == -1) {
    console.log("Not there in array");
    return true;

} else {
    console.log("Found it");
    return false;
}

edited Oct 24, 2016 at 16:56

answered Oct 24, 2016 at 15:24

Venkata Krishna

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1 Comment

aw04 Over a year ago

this is a fine alternative and all, but you don't even mention the problem with the code posted... this answer will likely only confuse the OP

Mamdouh Saeed · Accepted Answer · 2016-10-24 18:42:36Z

2

Update: using .indexOf() to detect if stat value is one of arr elements

Pure JavaScript

var arr = [20,30,40,50,60,70,80,90,100];
//or detect equal to all
//var arr = [10,10,10,10,10,10,10];
    var stat = 10;

if(arr.indexOf(stat)==-1)alert("stat is not equal to one more elements of array");

edited Oct 24, 2016 at 18:42

answered Oct 24, 2016 at 15:36

Mamdouh Saeed

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2 Comments

aw04 Over a year ago

this answer seems entirely overly comlex when the problem is a simple logic error

Mamdouh Saeed Over a year ago

Updated to be sample using indexOf()

four · Accepted Answer · 2016-10-25 07:48:30Z

1

The logical OR '||' automatically short circuits if it meets a true condition once.

false || false || true || false = true, stops at second condition.

On the other hand, the logical AND '&&' automatically short circuits if it meets a false condition once.

false && true && true && true = false, stops at first condition.

answered Oct 25, 2016 at 7:48

four

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