Jquery - How to get the style display attribute "none / block"

Question

Is there a way to get the style: display attribute which would have either none or block?

DIV :

<div id="ctl00_MainContentAreaPlaceHolder_cellPhone_input_msg_container" class="Error cellphone" style="display: block;">       <p class="cellphone" style="display: block;">Text</p>  </div> 

I know that there is a way to find out if the DIV is hidden or not but in my case this div is dynamically injected so it always shows up as visible false thus I cannot use that :

$j('.Error .cellphone').is(':hidden') 

I am able to get the result "display:block" using :

$j('div.contextualError.ckgcellphone').attr('style') 

Is there a way to get just the value "block" or "none" or is there a better/more efficient way to do this?

Answer 1

You could try:

$j('div.contextualError.ckgcellphone').css('display') 

Answer 2

If you're using jquery 1.6.2 you only need to code

$('#theid').css('display') 

for example:

if($('#theid').css('display') == 'none'){     $('#theid').show('slow');  } else {     $('#theid').hide('slow');  }