jQuery prependTo（）选择器

Question

I have a code like this:

 <div class="usa_img_bg"> <div class="ca_title">Lorem Ipsum</div> <div class="img_box"> <div class="photo_room_f"> <p>Lorem Ipsum dolor si amet...</p> </div> </div> </div> <div class="usa_img_bg"> <div class="ca_title">Lorem Ipsum</div> <div class="img_box"> <div class="photo_room_f"> <p>Lorem Ipsum dolor si amet...</p> </div> </div> </div> 

I want to get the ca_title div and move it into the photo_room_f div for every usa_img_bg .

I use this code:

 $.each($('.usa_img_bg'), function(index, element) { $(element).children(".ca_title").prependTo(".photo_room_f"); }); 

but it gets me all titles and puts them all on every usa_img_bg div. How can I make it get only the ca_title that's in every single usa_img_bg ?

Thanks in advance!

Answer 1

The selector you're using inside prependTo is a standard jquery selector and it doesn't know you're only interested in the current element , so it will be returning ALL elements on the page with a class of photo_room_f .

Try something like this:

$.each($('.usa_img_bg'), function(index, element) {
    var $el = $(element);

    $el.children(".ca_title").prependTo($el.find(".photo_room_f"));
});

Answer 2

The issue in your code is that you are appending to all the .photo_room_f elements, you need to find() the one related to the .usa_img_bg of the iteration. Try this:

 $('.usa_img_bg').each(function(i, e) { var $room = $(this).find(".photo_room_f"); $(this).find(".ca_title").appendTo($room); }); 

 <script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script> <div class="usa_img_bg"> <div class="ca_title">Lorem Ipsum</div> <div class="img_box"> <div class="photo_room_f"> <p>Lorem Ipsum dolor si amet...</p> </div> </div> </div> <div class="usa_img_bg"> <div class="ca_title">Lorem Ipsum</div> <div class="img_box"> <div class="photo_room_f"> <p>Lorem Ipsum dolor si amet...</p> </div> </div> </div> 

Answer 3

Try this:

$.each($('.usa_img_bg'), function() { 
    $(this).children(".ca_title").prependTo($(this).find('.photo_room_f'));
});

Answer 4

Try utilizing context parameter of jQuery()

.prependTo($(".photo_room_f", element));

where element is context : .usa_img_bg

 $.each($(".usa_img_bg"), function(index, element) { $(element).children(".ca_title").prependTo($(".photo_room_f", element)); }); 

 <script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js"> </script> <div class="usa_img_bg"> <div class="ca_title">Lorem Ipsum</div> <div class="img_box"> <div class="photo_room_f"> <p>Lorem Ipsum dolor si amet...</p> </div> </div> </div> <div class="usa_img_bg"> <div class="ca_title">Lorem Ipsum</div> <div class="img_box"> <div class="photo_room_f"> <p>Lorem Ipsum dolor si amet...</p> </div> </div> </div>