- if ( getComputedStyle && !color( elem ) )
- ret = getComputedStyle.getPropertyValue( name );
-
- // If the element isn't reporting its values properly in Safari
- // then some display: none elements are involved
- else {
- var swap = [], stack = [];
-
- // Locate all of the parent display: none elements
- for ( var a = elem; a && color(a); a = a.parentNode )
- stack.unshift(a);
-
- // Go through and make them visible, but in reverse
- // (It would be better if we knew the exact display type that they had)
- for ( var i = 0; i < stack.length; i++ )
- if ( color( stack[ i ] ) ) {
- swap[ i ] = stack[ i ].style.display;
- stack[ i ].style.display = "block";
- }
-
- // Since we flip the display style, we have to handle that
- // one special, otherwise get the value
- ret = name == "display" && swap[ stack.length - 1 ] != null ?
- "none" :
- document.defaultView.getComputedStyle( elem, null ).getPropertyValue( name ) || "";
-
- // Finally, revert the display styles back
- for ( var i = 0; i < swap.length; i++ )
- if ( swap[ i ] != null )
- stack[ i ].style.display = swap[ i ];
- }