I have this code:
$(window).scroll(function(){
$('#globalnav').css('left',-$(window).scrollLeft());
$('#globalborder').css('left',-$(window).scrollLeft());
$('#backgroundlayer').css('left',-$(window).scrollLeft());
});
How can I make al the variables compiled in just one line? So it should like something like this
$(window).scroll(function(){
$('#globalnav','#globalborder','#backgroundlayer').css('left',-$(window).scrollLeft());
});
Just drop some quotes
$(window).scroll(function(){
$('#globalnav, #globalborder, #backgroundlayer').css('left',-$(window).scrollLeft());
});
If you pass the elements as a comma separated list in one single string to the selector, they are all added.
If you use a comma separated list of different strings, you're using the "context selector", which is a shortcut for find() and works differently.
$/jQuery function will attempt to match the arguments following one of its signatures, as seen in the API docs. However, none of its signatures takes 3 arguments, and as some quick console testing via dev tools revealed on the jQuery website, an empty jQuery object is returned. Presumably, this is the default behavior if it cannot match arguments to one of its signatures.
$('.child', '#parent') is equal to $('#parent').find('.child'), of course trying to call that with three arguments fails.
,outside of the', you are passing multiple arguments to the jQuery$function. Using adeneo's answer, you are passing a single argument to$, which is a string representing a list of selectors.uploadplugin that questioner was using, rather than an issue selecting multiple items byid, despite the title.