Difference between PHP and JS evaluation of variables

Can someone please explain to me why the following javascript code produces an alert with 321 and the PHP code produces 1.

I kno开发者_如何学JAVAw the PHP code evaluates the expression and returns true or false. What I don't know is why in JavaScript it works like a ternary operator. Is it just the way things were implemented in the language?

var something = false;
var somethingelse = (something || 321);
alert(somethingelse); // alerts 321

$var = '123';
$other = ($var || 321);
echo $other; // prints 1

Thanks!

Is it just the way things were implemented in the language?

Yes, JavaScript does it a bit differently. The expression (something || 321) means if something is of a falsy value, a default value of 321 is used instead.

In conditional expressions || acts as a logical OR as usual, but in reality it performs the same coalescing operation. You can test this with the following:

if ((0 || 123) === true)
    alert('0 || 123 evaluates to a Boolean');
else
    alert('0 || 123 does not evaluate to a Boolean');

In PHP the || operator performs a logical OR and gives a Boolean result, nothing else.

I'm actually surprised javascript did not alert 1 or true as well. The syntax you want for js is:

var somethingelse = something || 321;

Wrapping parentheses around something evaluates it as truthy / falsey. For php your are saying:

//$other will equal true if $var is true or 321 is true. 
$other = ($var || 321);

A matching statement in php would look like:

$other = ($var) ? $var : 321;

Just to add on boltClock answer since I can't comment - If you want it be a Boolean value you can parse it to bool like this:

var somthing = !!(somthingelse || 321);

In PHP ($var || 321); is evaluated and assigned to $other.

You can use this in PHP.

($other = $var) || $other = 321;

Update: As BoltClock said in Javascript var somethingelse = (something || 321) seeks to assign a default value to the variable if something is false.