Visibility attribute question

Question

What is the difference between

$(\"#myid\").attr(\"style\", \"visibility: hidden\")

and

$(\"#myid\").css(\"visibility\", \"hidden\"

Answer 1

Nothing. Just two ways to accomplish the same goal.

The first will overwrite any existing style settings. If you had:

<div style="font-weight: bold;" />

It would become:

<div sytle="visibility: hidden;" />

The second will add the visibility setting to the existing styles. So:

<div style="font-weight: bold;" />

Woudl become:

<div style="font-weight: bold; visibility: hidden;" />

If there's no style attribute already set, then the two will produce the same end result.

Answer 2

Doing this:

$("#myid").attr("style", "visibility: hidden")

Will leave only this style attribute, while doing this:

$("#myid").css("visibility", "hidden")

Will add (or set) this style attribute.

Here's an example, the first will always result in this:

style="visibility: hidden;"

The second just adds visibility so your style may now be:

style="width: 50px; color: red; visibility: hidden;"

Answer 3

There isn't really any difference. $.css() is just a shortcut method for accessing the css style attribute of a dom element.

http://api.jquery.com/css/

EDIT: As justin pointed out, there is a difference in that the .attr() method will overwrite any existing style attributes.