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如何获得父div内元素的像素距离

[英]How to get distance in pixels of element inside of parent div

so I'm using Jquery and pure js to write small application, I have a element inside of a div parent.This Draggable element (ui widget) that can be moved around in div parent.I would like to get exact amount of pixels from the top and from the left. 所以我使用Jquery和纯js编写小型应用程序,我在div父级中有一个元素。这个Draggable元素(ui小部件)可以在div父级中移动。我想从中获取确切的像素数量顶部和左侧。

This is what I mean: 这就是我的意思: 在此处输入图片说明

I have following code right now, but with this code I can't get exact ammount of pixels from top and bottom, 我现在有以下代码,但是使用此代码,我无法从顶部和底部获得确切的像素数量,

$('#dragThis').draggable(
{
    drag: function(){
        var parentOffset =$(this).parent().offset(); 
        var offset = $(this).offset();
        var xPos = offset.left - parentOffset.left ;
        var yPos = offset.top - parentOffset.top  ;
        $('#posX').text('x: ' + xPos);
        $('#posY').text('y: ' + yPos);
    },
    containment: "#parentDiv"
});

Working example: http://jsfiddle.net/8fbdf93m/ 工作示例: http : //jsfiddle.net/8fbdf93m/

Please help. 请帮忙。

Try Jquery.Position method. 尝试使用Jquery.Position方法。 It allows you to retrieve the current position of an element relative to the offset parent 它允许您检索元素相对于偏移父级的当前位置

you can also try $(this).css('left') or $(this).css('top') 您也可以尝试$(this).css('left')$(this).css('top')

updated fiddle http://jsfiddle.net/8fbdf93m/4/ 更新小提琴http://jsfiddle.net/8fbdf93m/4/

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