> ... the fx .show/.hide methods use the same logic as
> the normal .show/.hide metheds (both .show methods
> set display 'block').
The non-fx show/hide try to preserve the original display value:
show: function(){
this.style.display = this.oldblock ? this.oldblock : "";
if ( jQuery.css(this,"display") == "none" )
this.style.display = "block";
},
hide: function(){
this.oldblock = this.oldblock || jQuery.css(this,"display");
if ( this.oldblock == "none" )
this.oldblock = "block";
this.style.display = "none";
},
There could be problems if someone used the fx hide and the standard show or
vice versa, since fx doesn't use oldblock.
I am not sure why .show even needs that extra check for display:none; how
could it be "none" when it was set on the line above and .hide() makes sure
it isn't "none" as well? Nothing else in jQuery seems to mess with oldblock.
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