CODEWITHSUNDEEP
javascript
Sky Red Tea
Sky Red Tea

Reputation: 29

Using multiple conditions within an if statement

How using multiple conditions within an if statement?

function testNum(a) {
  if (a == (1 || 2 || 3)) {
    return "is 1, 2, or 3";
  } else {
    return "is not 1, 2, or 3";
  }
}

console.log(testNum(1)); // returns "is 1, 2, or 3"
console.log(testNum(2)); // returns "is not 1, 2, or 3"
console.log(testNum(3)); // returns "is not 1, 2, or 3"

testNum(2) and testNum(3) should return: "is 1, 2 or 3" but doesn't.

Upvotes: 2

Views: 121

Answers (6)

Pranav C Balan
Pranav C Balan

Reputation: 115282

In this particular scenario, you can even use an array and Array#includes method for checking.

if ([1, 2, 3].includes(a)) {
  // your code
}


function testNum(a) {
  if ([1, 2, 3].includes(a)) {
    return "is 1, 2, or 3";
  } else {
    return "is not 1, 2, or 3";
  }
}
console.log(testNum(1));
console.log(testNum(2));
console.log(testNum(4));
console.log(testNum(3));
FYI : In your current code (1 || 2 || 3) results 1(since 1 is truthy) and actually a == (1 || 2 || 3) does a == 1. The right way is to seperate each conditions with || (or), for eg : a == 1 || a == 2 || a ==3.

For more details visit MDN documentation of Logical operators.

Upvotes: 8

Vishnu Mohan
Vishnu Mohan

Reputation: 121

The reason why you are not getting the expected result because you are performing a Logical OR operation here.

As per MDN Network : Logical OR (||) expr1 || expr2 If expr1 can be converted to true, returns expr1; else, returns expr2.

(1 || 2 || 3) is evaluated and the operation results at 1. You can check the result by pasting (1 || 2 || 3) in Chrome debugger or something.

So when the first number is 1 in all cases, the expression result is also 1. You can change the expression as how others have advised to get the desired output.

Upvotes: 0

Joe Iddon
Joe Iddon

Reputation: 20434

Your or-operators are placed incorrectly:

function testNum(a) {
    if (a == 1 || a == 2 || a == 3) {
        return "is 1, 2, or 3";
    } else {
        return "is not 1, 2, or 3";
    }
}

Before, you were testing if a was equal to 1 || 2 || 3 which evaluates to 1 †. So you were just checking a == 1 which is not what you wanted!

† Essentially when you string together "or"s like this, the first truthy value is returned. For example, you can assert for yourself that: 0 || False || 5 gives 5.

Upvotes: 4

mbojko
mbojko

Reputation: 14699

You can either

    if (a == 1 || a == 2 || a == 3) {

or

    if ([1, 2, 3].includes(a)) {

which will be more convenient when there's more values to test. If you want to be compatible with ancient browsers, you have to write it like this

    if ([1, 2, 3].indexOf(a) >= 0) {

which look undeniably uglier.

Upvotes: 0

Miljenko Barbir
Miljenko Barbir

Reputation: 1193

You can try using a list of qualified values and check if the lookup value belongs there:

function testNum(a) {
    var candidates = [1, 2, 3];
    if (candidates.indexOf(a) != -1) {
        return "is 1, 2, or 3";
    } else {
        return "is not 1, 2, or 3";
    }
}

console.log(testNum(1)); // returns "is 1, 2, or 3"
console.log(testNum(2)); // returns "is not 1, 2, or 3"
console.log(testNum(3));
console.log(testNum(4));

Upvotes: 0

Praveen Kumar Purushothaman
Praveen Kumar Purushothaman

Reputation: 167250

You cannot have || like that. The one you have used is not the right way. You should be using:

function testNum(a) {
  if (a == 1 || a == 2 || a == 3) {
    return "is 1, 2, or 3";
  } else {
    return "is not 1, 2, or 3";
  }
}

console.log(testNum(1));
console.log(testNum(2));
console.log(testNum(3));

Upvotes: 6

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