如何设置div位置依赖于另一个div位置？

Question

This is the html part of the code:

<div class="snake">
    <div class="body"></div>
</div>

The function below takes the last block, removes it and creates another one to put it at the top.

//The snake turns right
var turnRight = function() {
    blocks_moved++;
    $('.snake').children().last().remove();
    $('<div></div>').addClass('body').prependTo('.snake').css('top', 0 + '%').css('left', xPos * blocks_moved * 1.2 + '%');
    $('.snake').children().removeClass('head');
    $('.snake').children().first().addClass('head');
...more code

With the code below I tried to set the snake position eqal to head's position so every time a new block becomes the head the snake moves by one block.

//Set the snake position equal to head's position
var headPos = $('.snake').children().first().position();
    $('.snake').position(headPos);

        if (blocks_moved === length) {
            blocks_moved === 0;
        }
    }

When I call the function turnRight(); the head and body move but snake stays in place. I also tried with .css() method but I didn't succeed. Any other ways to do this?

I solved this problem with .offset() and positioning. Here is a live example: https://jsfiddle.net/dusth7gc/3/

Answer 1

solved this problem with .offset() and positioning.

var turnRight = function() {

    $('.body').css('background-color', 'gray');

    headPos_x = ($('.head').offset().left / $(window).width() * 100);
    headPos_y = ($('.head').offset().top / $(window).height() * 100);  

    $('<div></div>').addClass('body').prependTo('.snake').css('top', headPos_y + '%').css('left', headPos_x + helper_width * 1.2 + '%');

            $('.snake').children().last().remove();
            $('.snake').children().removeClass('head');
            $('.snake').children().first().addClass('head');



            $('.helper').css('top', $('.head').offset().top + 'px').css('left',$('.head').offset().left + 'px');

        if(collision($('.helper'), $('.meal')) === true) {
            $('.meal').remove();
            $('<div></div>').addClass('body').css('background-color', 'transparent').appendTo('.snake');
            foodCreate();
        }
}

Here is a live example: https://jsfiddle.net/dusth7gc/3/ .