jQuery: move window viewport to show freshly toggled element

问题

I have a snippet of jQuery in doc ready which toggles a div containing a textarea:

$('div#addnote-area').hide(); // hide the div
$('a#addnote-link').click(function() { // click event listener on link
     $('div#addnote-area').toggle(); // toggle the hidden div 
});

The toggle works fine when clicking the link. The problem I'm having is that if div#addnote-area is below the browser's current viewport, it remains there when it's shown. I'd like the user's cursor to go to the textarea and for the whole textarea to be viewable in the window.

回答1:

Check out the scrollTo jQuery plugin. You can simply do something like this:

$.scrollTo('div#addnote-area');

Or, if you want to animate it, provide the # of milliseconds for the scrolling to last:

$.scrollTo('div#addnote-area', 500);

回答2:

Without the scrollTo plugin

$(window).scrollTop($('div#addnote-area').offset().top)

EDIT: Well I sure get a lot of points from this old answer :)

Here's a bonus, this can also be animated.

Just use the animate() function and target the body tag:

$('body').animate({scrollTop:$('div#addnote-area').offset().top},500)

Try it on Stackoverflow! Open the inspector console and run

$('body').animate({scrollTop:$('#footer').offset().top},500)

回答3:

jQuery's scrollTop also works. Try setting like:

 $('#idOfElement').css('scrollTop', 10)

You can get height/width pretty easily using $(...).offset().

来源：https://stackoverflow.com/questions/2329281/jquery-move-window-viewport-to-show-freshly-toggled-element