Dave Goodchild wrote:
I use a config file too. That was a sanity check. The file extract
looked like this:
$months = array(1 => 'January', 2 => 'February', 3 => 'March', 4 =>
'April', 5 => 'May', 6=> 'June',
7 => 'July', 8 => 'August', 9 => 'September', 10 =>
'October', 11 => 'November', 12 => 'December');
which was called in with require_once. The reference to $months in the
calling page, checked with var_dump, was NULL.
When I wrapped it like this:
function getmonths() {
$months = array(1 => 'January', 2 => 'February', 3 => 'March', 4
=> 'April', 5 => 'May', 6=> 'June',
7 => 'July', 8 => 'August', 9 => 'September', 10 =>
'October', 11 => 'November', 12 => 'December');
return $months;
}
it worked. Not sure why the simple variable didn't work.
On 14/08/06, *Adam Zey* <azey@xxxxxx <mailto:azey@xxxxxx>> wrote:
Dave Goodchild wrote:
> Hi all - I have several require_once statements in my web app to
load in
> small function libraries. A common one bundles a variety of
functions to
> handle date math and map month numbers to month names. I
originally defined
> an array in that file plus a bunch of functions but when I
loaded the page,
> the array variable, referenced further down the page, was NULL.
I wrapped a
> function def around the array and returned it and all was fine.
>
> I may be suffering from mild hallucinations, but can you not define
> variables in a required file? It is not a scope issue as the
array variable
> is referenced in the web page, not in any function.
>
I know for a fact that you can define variables in PHP 4 and 5.
The idea
behind include and require is little more complex than copying and
pasting the code. Many of my scripts include a config.php which has
various variables created with setting information.
Regards, Adam Zey.
--
http://www.web-buddha.co.uk
http://www.projectkarma.co.uk
Was the $months variable created inside an if statement or something
else? PHP's rules of scope say that variables created inside a code
block (like an if, a for, a while, a foreach), they stop existing the
moment you exit that code block. So this:
$foo = "bar";
if ( $foo == "bar" )
{
$baz = "narf";
}
echo $baz;
That code will output nothing, because $baz is empty by the time I try
to output it. The solution that I use is this:
$foo = "bar";
$baz = "";
if ( $foo == "bar" )
{
$baz = "narf";
}
echo $baz;
In which case the output would be "narf", because the variable existed
before I changed it in the if. This sounds like it might be your
problem, though I can't know without seeing the code.
Regards, Adam Zey.
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