Question

Similar to: Using fadein and append

But the solutions there aren’t working for me. I’m trying:

 $('#thumbnails').append('<li><img src="/photos/t/'+data.filename+'"/></li>').hide().fadeIn(2000);

But then the whole list fades in at once, not as each item is added. It looks like hide() and fadeIn() are being applied to $('#thumbnails') not the <li>. How would I get them to apply to that instead? This doesn’t work either:

$('#thumbnails').append('<li stle="display:none"><img src="/photos/t/'+data.filename+'"/></li>').filter(':last').fadeIn(2000);

Other suggestions?

Answer 1

Your first attempt is very close, but remember that append() is returning #thumbnails, not the item you just added to it. Instead, construct your item first and apply the hide().fadeIn() before adding it:

$('#thumbnails')
    .append($('<li><img src="/photos/t/'+data.filename+'"/></li>')
        .hide()
        .fadeIn(2000)
    );

This uses the dollar function to construct the <li> ahead of time. You could also write it on two lines, of course, if that makes it clearer:

var item = $('<li><img src="/photos/t/'+data.filename+'"/></li>')
    .hide()
    .fadeIn(2000);
$('#thumbnails').append(item);

Edit: Your second attempt is also almost there, but you need to use children() instead of filter(). The latter only removes nodes from the current query; your newly-added item isn’t in that query, but is a child node instead.

$('#thumbnails')
    .append('<li style="display:none"><img src="/photos/t/'+data.filename+'"/></li>')
    .children(':last')
    .hide()
    .fadeIn(2000);