如何从jquery中的父元素获取子元素的xy坐标? [英] How to get xy coordinates of child element from parent element in jquery?

问题描述

如何从父元素而不是(body)获取子元素的xy坐标?如何从DIV B到DIV A获取x值，所以当窗口调整大小时，值保持为真??

How to get xy coordinates of child element from parent element, and not the(body)? How to get x value from DIV B to DIV A, so when window resizes, the values stay true???

var startPos = $("#divB").position();
$("#divB").draggable({ 
     containment: 'parent',
     stop: function(event, ui) {
        var stopPos = $(this).position();
        $("#firstInput").val((stopPos.left - startPos.left));
     }
});

是否有一种方法可以直接获取此值，而不是从整个屏幕的xy坐标中获取?

Is there a method to get this values directly and not from xy coordinates of whole screen?

推荐答案

要获取子元素相对于其父元素的偏移量，请使用position():

To get the offset of a child element relative to its' parent use position():

$("#child").position().left;
$("#child").position().top;

对此的警告是，父母必须不必须为position: static才能起作用.如果父对象是静态的，则jQuery将仅返回offset()-我认为这是您的代码中发生的事情.

The caveat on this being that the parent must NOT be position: static for this to work. If the parent is static, jQuery will just return the offset() - which I assume is what is happening in your code.

static是大多数块元素的默认设置，因此您需要在CSS的父级上将position设置为fixed，absolute或relative.

static is the default for most block elements, so you'll need to set position to fixed, absolute or relative on the parent in your CSS.

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