Really basic javascript, checking a condition using || (or)

Question

I am very new to JavaScript, and trying to learn using lynda.com.

I'm currently playing around with the if statement.

The code in question:

var value = prompt("1 or 2?");

if ( value == 1 ) {
    alert("You have chosen 1.");
} 

if ( value == 2 ) {
    alert("You have chosen 2.");
}   

if ( value != (1 || 2) ) {
    alert("Invalid number. Try again.");
}

This is what i want to happen: If the number is NOT 1 or 2, i want JS to execute this piece of code:

 if ( value != (1 || 2) ) {
    alert("Invalid number. Try again.");
}

Apparently this is not the correct way to write it out, and i've tried writing it a bunch of different ways. Can someone show me the correct way, possibly using the else statement?

Though I don't use CoffeeScript myself, you can do something like this: if value not in [1, 2] then if you ever decide to venture into the preprocessor route some day. — Jay
– Jay, Commented Aug 15, 2013 at 13:18

tobspr · Accepted Answer · 2013-08-19 22:20:10Z

4

if ( value != 1 && value != 2) {
    alert("Invalid number. Try again.");
}

If you write value != (1||2), the (1||2) will get first evaluated to 1 and then compared to value, so you effectively wrote:

if ( value != 1 ) { [...] }.

This is because the () have a higher predence than !=. You can also read this explanation about operator predence which gives more details.

edited Aug 19, 2013 at 22:20

answered Aug 15, 2013 at 13:10

tobspr

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4 Comments

Adam Jenkins Over a year ago

I believe the OP effectively wrote if(value != 1)

Bergi Over a year ago

No. It would get evaluated to 1.

tobspr Over a year ago

Press F12 and enter (1||2) - it will return 1

Adam Jenkins Over a year ago

Yes, @TobSpr - do that and you'll see the result is 1

Vitaly · Accepted Answer · 2013-08-15 13:13:17Z

3

1

if ( value != 1 && value !=2 ) {
    alert("Invalid number. Try again.");
}

2

 if ( !(value == 1 || value ==2) ) {
    alert("Invalid number. Try again.");
}

3

if ( value == 1 ) {
    alert("You have chosen 1.");
} 
else if ( value == 2 ) {
    alert("You have chosen 2.");
}   
else {
    alert("Invalid number. Try again.");
}

answered Aug 15, 2013 at 13:13

Vitaly

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Comments

mkuk · Accepted Answer · 2013-08-15 13:14:59Z

2

The best way to write this statement would be as follow:

if ( value == 1 )
{
  alert("1");
}
else if ( value == 2 )
{
  alert("2");
}
else
{
  alert("no 1 or 2");
}

The if statement that you are messing on is (1 || 2) What will happen is it will do a Boolean test and return 1. it should look like this

if ( value !== 1 && value !== 2 )
{
  alert("no 1 or 2");
}

Thanks

answered Aug 15, 2013 at 13:14

mkuk

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Comments

msturdy · Accepted Answer · 2013-08-15 13:16:53Z

2

As already noted, the best is to separate into two NOT statements,and evaluate both:

if ( value != 1 && value != 2) ) {
    alert("Invalid number. Try again.");
}

However, you could also use the if, else if, else pattern to cover yourself against all other inputs (letters, punctuation, whitespace, etc). The else acts as a catch-all at the end:

var value = prompt("1 or 2?");

if ( value == 1 ) {
    alert("You have chosen 1.");
} 

else if ( value == 2 ) {
    alert("You have chosen 2.");
}   

else {
    alert("Invalid number. Try again.");
}

answered Aug 15, 2013 at 13:16

msturdy

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Comments

Adam Jenkins · Accepted Answer · 2013-08-15 13:13:09Z

1

Inline statements like: (1 || 2) only evaluate the right hand side of the || if the left hand side is false-y. So what your statement actually is saying is:

if ( value != 1 )

Because 1 evaluates to true.

As most others have pointed out, what you actually want to do is:

if ( value != 1 && value != 2 )

answered Aug 15, 2013 at 13:13

Adam Jenkins

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Comments

Oscerd · Accepted Answer · 2013-08-15 13:14:29Z

1

The logic is incorrect. You have to do in this way

if ( value != 1 && value != 2) ) {
    alert("Invalid number. Try again.");
}

the (1||2) statement is evaluated as 1 so you are testing if value != 1

answered Aug 15, 2013 at 13:14

Oscerd

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Comments

iConnor · Accepted Answer · 2013-08-15 13:21:57Z

1

have tried writing

if(value != 1 && value != 2)

edited Aug 15, 2013 at 13:21

iConnor

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answered Aug 15, 2013 at 13:11

karan421

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4 Comments

Bergi Over a year ago

Tell me a value that doesn't pass this condition.

Saint Over a year ago

the value would be 1 and the condition is not meet

Bergi Over a year ago

@Saint: He's changed || to && now.

Saint Over a year ago

did not notice... sorry

mavili · Accepted Answer · 2013-08-15 13:11:18Z

0

You have to do it like

if ( value != 1 && value != 2 ) {
    alert("Invalid number. Try again.");
}

OR

if (!(value == 1 || value == 2)) {
    alert("Invalid number. Try again.");
}

answered Aug 15, 2013 at 13:11

mavili

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Comments

Hoa · Accepted Answer · 2013-08-15 13:21:00Z

0

You can use switch as follows

switch(value) {
  case 1:
    alert("You have chosen 1.");
    break;
  case 2:
    alert("You have chosen 2.");
    break;
  default:
    alert("Invalid number. Try again.");
}

answered Aug 15, 2013 at 13:21

Hoa

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Comments

iConnor · Accepted Answer · 2013-08-15 13:24:18Z

0

Your problem is here:

if ( value != (1 || 2) )

What ths is doing is checking if "value" is not equal to (1 OR 2), where (1 OR 2) is evaluated first, because it's in the brackets. (1 OR 2) evaluates to 1, since 1 is truthy in javascript. What your code is actually doing is checking if your value is not equal to 1.

What you actually want to do is check if your value is !=1 and that your value is !=2. You do that like this:

if (value !=1 && value !=2) { /* do something */ }

edited Aug 15, 2013 at 13:24

iConnor

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answered Aug 15, 2013 at 13:14

Racheet

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1 Comment

iConnor Over a year ago

No problem, just fixed your SyntaxError: Unexpected end of input

gen_Eric · Accepted Answer · 2013-08-15 13:24:33Z

-1

You don't need a clause. You can just use:

else{
}

edited Aug 15, 2013 at 13:24

gen_Eric

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answered Aug 15, 2013 at 13:18

mutant_america

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