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.
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));(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.
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));
|| like that, it just does not do what the OP assumed it did. I think you should clarify what it WAS doing.
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.
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));
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.
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.
