如何使用JQuery仅删除Style attr的位置（顶部和左侧）？

Question

I am dynamically setting the position attr of the element only when the view is out of viewport of the window, in other case default value is set from the css file ,

.css( { "left": (left + 20) + "px", "top": (top+10) + "px" } );

Once the dynamic position is set, I want to remove the position attr alone.

I can remove style attribute it will also my display property of style which is required.

Is there a way to to remove the position attr alone?

Answer 1

Perhaps, your best Bet would be to simply put them to their default value. Top and left have the default value "auto". So:

jQuery(selector).css({
   'top': 'auto',
   'left': 'auto'
})

Answer 2

$(el).css('top', ''); will remove the inline style declaration, reverting back to either what is specified in your stylesheet or the default.

Answer 3

I'm not sure if you're asking for removing a position HTML attribute, a position style, or top and left styles. In any case, you can remove any attribute with $.removeAttr() , and you can remove (dynamically assigned) styles with $.css({ styleName: '' }) ;

Edit: it seems you want $.css({ position: '' }) , or more likely $.css({ position: 'static' }) , as that is the default value.