I think there's some issues with fadeIn/fadeOut for <tr>.
Try something like the following:
var html = '<tr id="ohai"><td class="hide">hello</td></tr>';
$('#groups').append(html).find("td").fadeIn("fast");
The class "hide" on the <td> is style with display:none;
Add this to all your <td> if you have more than one.
On Jul 13, 2:05 pm, Nate <nathanle...@gmail.com> wrote:
> LOL - I swear I thought of almost the exact same thing when I went out
> and took a drive...it's amazing what time away from the computer
> does...and of course, a little help.
>
> Now I seem to be missing something (yeah, I am new to jQuery...)
>
> I still want that table row to fade in...but neither method works:
>
> $('#groups').append('<tr id="ohai"><td>hello</td></tr>').fadeIn
> ("fast");
>
> $('#groups').fadeIn("fast").append('<tr id="ohai"><td>hello</td></
> tr>');
>
> On Jul 13, 7:35 pm, James <james.gp....@gmail.com> wrote:
>
> > How about just removing the empty <tr></tr> and use something like the
> > following for your insert row after your AJAX response:
>
> > $("#groups").append('<tr id="group'+data.id+'"><td>....</td></tr>');
>
> > On Jul 13, 1:00 pm, Nate <nathanle...@gmail.com> wrote:
>
> > > Also, James, each table row has a unique id. it will be called
> > > something like id="group34" or id="group35" - no two are the same.
> > > It's info that is related to the ID of the database.
>
> > > So when my jQuery posts to addgroup.php, it inserts the data into the
> > > database, and then I use the mysql_inser_id() function to retrieve the
> > > ID of the entry I just made. I then return that data to the main
> > > page, and I simply want to stick it below all of the data that was
> > > generated by PHP.
>
> > > To give you an idea - here is a screen capture of it in action, and
> > > some of the issues along with it:
>
> > >http://www.nathanledet.com/removingdata_jquery.mov
>
>
