Append and Slide together jQuery

China☆狼群 提交于 2019-12-02 22:02:39

Like SimpleCoder's solution, but in only one line using appendTo():

$('<div style="display: none;" class="new-link" name="link[]"><input type="text" /></div>').appendTo($('.insert-links')).slideDown("fast");

Demo: http://jsfiddle.net/V4SVt/336/

append() returns a reference to the original selector, not what was appended. I think you are looking for this:

$('.insert-links').append('<div style="display: none;" class="new-link" name="link[]"><input type="text" /></div>')
$('.insert-links').find(".new-link:last").slideDown("fast");

Live demo:

http://jsfiddle.net/V4SVt/2/

Although SimpleCoder's solution is perfectly valid, I'd do it like this:

i = 0;
$('#add-link').click(function() {   
    if(i < 9) {
        $('.insert-links').append('<div style="display: none;" class="new-link link_' + i + '" name="link[]"><input type="text" /></div>');
        $('.link_' + i).slideDown("fast");
        i++;
    }
    if(i == 9) {
        $('#add-link').fadeOut('200');
    }
});

The reason for it would be that each link input would get an ID in the form of a second class, which would come very handy for selecting them in case one wants to remove an element, should one accidentally add more than one needs. I have also added a fade out effect on the add-link element so the user doesn't get confused that it just disappeared. Of course, it is just a matter of personal taste, but I find it more user-friendly.

Hope this helps.

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