Question

function ajaxFunction2(){
var chooser2 = $('#chooser2').val();
$('#floatcontainertop').css("background",$('#chooser2').val());
}

Could someone please tell me how to fix my above code so that it works. I have absolutely no knowledge of jquery and I do not know how to call a variable within a function that I have already predefined. Thanks for the help!

Chooser2 represents a color, i.e. red, blue, yellow, black.

Answer 1

function ajaxFunction2(){
   var chooser2 = $('#chooser2').val();
   $('#floatcontainertop').css("background",chooser2);
}

IF all you need to do is use the variable, then that would be the way, however, if this isn’t changing it, it might be that there’s another, more subtle, issue?

What behaviour do you get right now?