Button of widget doesn't change immage

[Mirco]

Ok, thanks!
How can I see if a widget is open/close in JavaScript?

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I have resolve my previous question, but now I am not able to change the image of an object with JavaScript, do you know how to do it?
I tryed some istructions but no one worked..

This is my code:

Code:

$(function()
{
 var widgetS = false, widgetF = false;
 
 function openWidget(id_icon)
 {
   $(id_icon).css("background-image", "/scada/resources/icons/Button_Preessed.svg");  //This doesn't work
 }
 
 function closeWidget(id_icon)
 {
   $(id_icon).css("background-image", "/scada/resources/icons/empty.svg");  //This doesn't work
 }
 
 
 $(".widgetS").on("click", function()
 {
     if (widgetS == false)
   {
     openWidget(".widgetS");
     widgetS = true;
   }
   else
   {
     closeWidget(".widgetS");
     widgetS = false;
   }
   
   if (widgetF == true)
   {
     closeWidget(".widgetF");
     widgetF = false;
   }
 });
 
 $(".widgetF").on("click", function()
 {
     if (widgetF == false)
   {
     openWidget(".widgetF");
     widgetF = true;
   }
   else
   {
     closeWidget(".widgetF");
     widgetF = false;
   }
   
   if (widgetS == true)
   {
     closeWidget(".widgetS");
     widgetS = false;
   }
 });
 
 $(".pageS").on("click", function()
 {
     if (widgetS == true)
   {
     closeWidget(".widgetS");
     widgetS = false;
   }
   
   if (widgetF == true)
   {
     closeWidget(".widgetF");
     widgetF = false;
   }
 });
 
 $(".pageF").on("click", function()
 {
     if (widgetS == true)
   {
     closeWidget(".widgetS");
     widgetS = false;
   }
   
   if (widgetF == true)
   {
     closeWidget(".widgetF");
     widgetF = false;
   }
 });
});