jQuery在console.log中工作，但在变量中工作

Question

I have this console.log()

console.log($($('.biggie img:visible')[0]).attr('src').split('/').pop());

This returns:

4.jpg

When i put the line into a variable like this:

var jpg = $($('.biggie img:visible')[0]).attr('src').split('/').pop(); 

The console says:

Uncaught TypeError: Cannot call method 'split' of undefined 

What is going on?

Thank you!!

Edit:

Sorry, sorry, I know what is going on now, you pointed me all into the right direction below my answer...here is the whole code:

imgVis.fadeOut(400, function() {
                var jpg = $($('.biggie img:visible')[0]).attr('src').split('/').pop(),
                    nr = parseInt(jpg,10);
                $(this).next(':hidden').fadeIn(400);
                console.log($($('.biggie img:visible')[0]).attr('src').split('/').pop());
});

I actually had no image visible. Ashaming. :/ Thank you!!!

Answer 1

Hm ... this is really strange. Are these two lines one after each other. If not, then I'll suggest that your selector is not matching what you expect or there is no src attribute of the matched elements. I'll suggest two things:

1) Replace this

$($('.biggie img:visible')[0])

with just

$('.biggie img:visible')

What you are doing there is to convert the jQuery element to a DOM element and the again make it jQuery element.

2) console.log($('.biggie img:visible').length);

If the length is 0 , then there are still no visible images.