如何根据光标单击的位置放置弹出式div [英] How to position a popup div based on the position of where the cursor clicks

问题描述

我正在尝试在下面放置一个弹出div:

I am trying to position a popup div below:

<div style="display:none;height:200px;width:200px;border:3px solid green;" id="popup">Hi</div>

基于另一个div的点击.

based on the clicking of another div.

我正在.ready文件上运行此

I am running this on document .ready

$('div#d').bind('click', function (event) {
var offset = $(this).offset();
$('#popup').css('left',offset.left);    
$('#popup').css('top',offset.top);
$('#popup').css('display','inline');        
});

但是上面的内容甚至不会显示div

but the above will not even display the div

推荐答案

问题在于，offset()不能返回正确的鼠标位置，请尝试使用event.pageX和event.pageY:

The problem lie in the fact that offset() do not return the correct mouse position, try event.pageX and event.pageY instead:

$(document).ready(function(){
    $('div#d').bind('click', function (event) {
    $('#popup').css('left',event.pageX);      // <<< use pageX and pageY
    $('#popup').css('top',event.pageY);
    $('#popup').css('display','inline');     
    $("#popup").css("position", "absolute");  // <<< also make it absolute!
    });
});

请参见此处.

这篇关于如何根据光标单击的位置放置弹出式div的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！