如何从jquery中的父元素获取子元素的xy坐标?

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?

答

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.

如何从jquery中的父元素获取子元素的xy坐标?

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