Why is there no comparison statement in this javascript 'If… Else…' statement

问题

In the following code

var $next =  $active.next().length ? $active.next()
        : $('#slideshow IMG:first');

the part '$active.next().length' doesn't seem to compare anything and I don't understand how the condition is determined to be True or False.

Or is it saying that: if the various $next is equal to $active.next().length then the condition is true?

回答1:

In javascript any expression can be converted to a truthy or falsy value and hence is valid in a comparison place. The values which are false in javascript are

• false
• 0
• "" (empty string)
• null
• undefined
• NaN

In this case length refers to a numeric value and if it evaluates to 0 then it will be considered falsy. Otherwise it will be truthy

回答2:

If the length property is equal to 0 or undefined (i.e. $active is not an array), the condition will be false.

回答3:

If $active.next().length is true, which means that there is a next element, then $next = $active.next(). Otherwise $next = $('#slideshow IMG:first'). The ? operator is called the ternary operator. It is a short if else.

回答4:

It's a ternary comparison equivalent to:

if($active.next().length) {
    $next = $active.next();
}
else {
    $next = $('#slideshow IMG:first');
}

So the condition is based on $active.next().length which should return a value of zero or greater. Anything greater than zero, JavaScript will interpret as true, zero false.

回答5:

What you're looking at a Ternary opertation which is a short hand for If... Else... has you mentioned in the title.

So the long version of your statement is;

if($active.next().length){
 $next = $active.next();
}else {
 $next = $('#slideshow IMG:first');
}

来源：https://stackoverflow.com/questions/10401297/why-is-there-no-comparison-statement-in-this-javascript-if-else-statemen