如何使用JQuery获取CSS属性的值？

I am trying to set a box A to jump to the same x-position as box B. How do I do this? I tried the following but it does not work (the second line):

我试图设置一个方框A跳到与方框B相同的x位置。我该怎么做?我尝试了以下但它不起作用(第二行):

$("#boxa").animate({
   left: $("#boxb").left
}, 1000, function() { });
});

Thanks,

Sam

2 个解决方案

#1

either

$('#boxB').css('left');

or

$('#boxB').offset().left;

#2

For properties specified in pixels, using:

对于以像素指定的属性,使用:

$('#boxB').css('padding-left');

gives a string like: '22px'

给出一个字符串:'22px'

Maybe it's better to use:

也许最好使用:

parseInt($('#boxB').css('padding-left'), 10);

in such cases, which gives a number like 22 .

在这种情况下,给出一个像22的数字。