variable undefined

[jackvull]

I get a DJDiv is undefined error with the following:
function brieflyHideMenus(varDJID) {
var DJDiv = 'DJdiv' + varDJID;
alert(DJDiv);
document.getElementById(DJDiv).style.visibility='hidden';

setTimeout("document.getElementById(DJDiv).style.visibility='visible'", 5000)
}

It works up to the visibility.hidden part but then as soon as the stuff starts to run after 5 seconds (timout), the error appears.

Any ideas?
Thanks.

[kaled]

Code executed by setTimeout() is global in scope - it does not execute within the function. i.e. unless there is a global variable DJDiv that can act as a parameter the code will fail.

Kaled.

[jackvull]

Thanks that did it.
Variable scope annoys the **** out of me.

[Bernard Marx]

kaled is right. Yet I don't believe he was actually suggesting you should use a global variable for this purpose - he was just saying that that is what the command is expecting
- Am I right, kaled?

There are two solutions to this that don't require a global variable.
The second is my preferred, because it does away with string nonsense in the timer.

function brieflyHideMenus(varDJID) 
{ 
  var id = 'DJdiv' + varDJID;  
  document.getElementById(id).style.visibility='hidden';
  // note single quotes on id
  setTimeout("document.getElementById('"+id+'").style.visibility='visible'", 5000); 
} 
function brieflyHideMenus(varDJID) 
{ 
  var div = document.getElementById('DJdiv' + varDJID);
  div.style.visibility='hidden';
  setTimeout( function(){div.style.visibility='visible'}, 5000); 
} 

[kaled]

As Bernard said, I was explaining the cause rather than suggesting a solution.

As a general rule, when answering questions, I tend to take the "homework" approach. That is so say, I try to point people in the right direction rather than solve the problem.

Kaled.