Description:
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I have a column value coming back from a mysql query that is '%$g_userName'

Now, when I look at the column for each row and see a leading '%', I know that the value is actually the name of a variable that exists in my application.  Here's a code snippet, where $name contains the column's value as returned by mysql (the value is '%$g_userName'):

if (substr($name, 0, 1) == '%') {
   $actualName = substr($name, 1, strlen($name) - 1);
   $name = $$actualName;
}

When the code above gets executed, I get the following error:

Notice: Undefined variable: $g_userName in C:\htdocs\ebbs\scripts\rightMenu_inc.php on line 51


I added the following immediately after the if {...} to make sure $g_userName actually exists and contains a value:

echo $g_userName.'<br />';

And now I get the following:

Notice: Undefined variable: $g_userName in C:\htdocs\ebbs\scripts\rightMenu_inc.php on line 51
dave

Where 'dave' is the value I had hard-coded in $g_userName.



Reproduce code:
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$g_userName = 'dave';
$name = '$'.'g_userName';
$trueName = $$name;

echo $trueName;


Expected result:
----------------
dave

Actual result:
--------------
Notice: Undefined variable: $g_userName in C:\htdocs\ebbs\scripts\pageFooter_inc.php on line 7