My initial reaction is that your information might be more suited to a table, in which case you could use the tablesorter plugin.
http://tablesorter.com/docs/ On Nov 21, 5:58 pm, "javam.org" <alt...@gmail.com> wrote: > Hi there, > > My xhtml structure looks like: > > <div id="aaa"> > <div class="menu"> > <span class="s1">Name</span> > <span class="s2">Remaining Time</span> > <span class="s3">Status</span> > </div> > <div class="values"> > <ul id ="list"> > <li> > <span class="s1">Name value 1</span> > <span class="s2">Time value 1</span> > <span class="s3">Status value 1</span> > </li> > <li> > <span class="s1">Name value 1</span> > <span class="s2">Time value 2</span> > <span class="s3">Status value 3</span> > </li> > <li> > <span class="s1">Name value 3</span> > <span class="s2">Time value 3</span> > <span class="s3">Status value 3</span> > </li> > </ul> > </div> > </div> > > I am trying to sort span values for Name, Time and Status.. > I can implement to just only one menu item by adding <div ="1">, <div > ="2"> ... which covers li elments. But it cannot provide my needs. I > have to implement it to all. > > When I click time menu element (let assume "Time value 2" is greater > than "Time value 3" and "Time value 3" is greater than "Time value > 1"), all s2 spans should be at the top of the list... looks like: > > <div id="aaa"> > <div class="menu"> > <span class="s1">Name</span> > <span class="s2">Remaining Time</span> > <span class="s3">Status</span> > </div> > <div class="values"> > <ul id ="list"> > <li> > <span class="s2">Name value 1</span> > <span class="s3">Time value 1</span> > <span class="s1">Status value 1</span> > </li> > <li> > <span class="s2">Name value 1</span> > <span class="s3">Time value 2</span> > <span class="s1">Status value 3</span> > </li> > <li> > <span class="s2">Name value 3</span> > <span class="s3">Time value 3</span> > <span class="s1">Status value 3</span> > </li> > </ul> > </div> > </div> > > And should be available for other menu elements also... > > Could you show me the ways to do it? > > Best Regards.
