Variable interplay between include files and parent files?

Report · May 09, 2007

How can I define a variable in a parent file which is then used in some
logic in an include file? For example, if I have -

<?php

$foo = 1;

?>
....
<?php require_once('foo.php'); ?>

and in foo.php, I have -

<?php if ($foo == 1) { echo "FOO"; } ?>

it doesn't seem to work. Am I testing improperly, or what?

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Report · May 09, 2007

On Wed, 9 May 2007 08:21:24 -0400, "Murray *ACE*"
<forums@HAHAgreat-web-sights.com> wrote:

>How can I define a variable in a parent file which is then used in some
>logic in an include file? For example, if I have -
<snip>
>it doesn't seem to work. Am I testing improperly, or what?

It would seem so. It should work. The only thing that comes to mind is
possibly a scope problem if the variable is defined inside a function
and you're echoing output outside the function, or you are echoing the
output inside a function for a variable that was defined outside the
function.

Gary

Report · May 09, 2007

For me it always works with both "require_once" and "include" when introducing the to-be-passed variable immediately before the "include":

<?php
$foo = 1;
include (“foo.php”); // or "require_once"
?>

Report · May 09, 2007

Hmmm - I thought it was problematic having ANYTHING within the include
statement's block.

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"geschenk" <webforumsuser@macromedia.com> wrote in message
news:f1sqav$8vk$1@forums.macromedia.com...
> For me it always works with both "require_once" and "include" when
> introducing
> the to-be-passed variable immediately before the "include":
>
> <?php
> $foo = 1;
> include (?foo.php?); // or "require_once"
> ?>
>

Report · May 09, 2007

>>
Hmmm - I thought it was problematic having ANYTHING within the include statement's block.
>>

no why, the $foo = 1; variable is separated from the include statement and just preceeds it

Report · May 09, 2007

On Wed, 9 May 2007 15:45:35 +0000 (UTC), "geschenk"
<webforumsuser@macromedia.com> wrote:

>For me it always works with both "require_once" and "include" when introducing
>the to-be-passed variable immediately before the "include":

Really doesn't matter whether it is immediately before or 10,000 lines
before. If it's in scope, it should be just fine.

Gary

Report · May 09, 2007

That's what I thought, too. But I solved the problem by moving the PHP to
the include file!

--
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"Gary White" <reply@newsgroup.please> wrote in message
news:qvu343hed44c0i68bca39jd4i6na8oolt6@4ax.com...
> On Wed, 9 May 2007 15:45:35 +0000 (UTC), "geschenk"
> <webforumsuser@macromedia.com> wrote:
>
>>For me it always works with both "require_once" and "include" when
>>introducing
>>the to-be-passed variable immediately before the "include":
>
> Really doesn't matter whether it is immediately before or 10,000 lines
> before. If it's in scope, it should be just fine.
>
> Gary

Report · May 09, 2007

>>
That's what I thought, too. But I solved the problem by moving the PHP to the include file!
>>

huh ? 😉 makes me wonder what you did "move" -- the $foo variable ?

Report · May 09, 2007

Actually, I changed everything - just to keep you offbalance!

The routine is now dependent on the parent filename, not some arbitrary
variable. 8)

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"geschenk" <webforumsuser@macromedia.com> wrote in message
news:f1tdsu$38n$1@forums.macromedia.com...
>>>
> That's what I thought, too. But I solved the problem by moving the PHP to
> the include file!
>>>
>
> huh ? 😉 makes me wonder what you did "move" -- the $foo variable ?

Report · May 09, 2007

>>
Actually, I changed everything - just to keep you offbalance!
>>

LOL, thanks a lot ;-)

>>
The routine is now dependent on the parent filename, not some arbitrary variable. 8)
>>

good on ya, and that´s surely an appropriate solution now -- but what will you do when you *really* need to pass a variable to an included page ? ;-)

if you want, just post the code of what you initially had in mind, maybe there´s something wrong - or was it just that simple foo example ?

Report · May 09, 2007

It was exactly the foo example. My error was an undefined variable error.

--
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"geschenk" <webforumsuser@macromedia.com> wrote in message
news:f1ths4$7rh$1@forums.macromedia.com...
>>>
> Actually, I changed everything - just to keep you offbalance!
> >>
>
> LOL, thanks a lot ;-)
>
> >>
> The routine is now dependent on the parent filename, not some arbitrary
> variable. 8)
> >>
>
> good on ya, and that´s surely an appropriate solution now -- but what will
> you
> do when you *really* need to pass a variable to an included page ? ;-)
>
> if you want, just post the code of what you initially had in mind, maybe
> there´s something wrong - or was it just that simple foo example ?
>

Report · May 09, 2007

On Wed, 9 May 2007 14:59:44 -0400, "Murray *ACE*"
<forums@HAHAgreat-web-sights.com> wrote:

>That's what I thought, too. But I solved the problem by moving the PHP to
>the include file!

Curious. That sounds like one of those solutions that shouldn't work,
but the problem goes away when you do it.

Gary

Report · May 09, 2007

Yeah - of course, I changed everything about it, too!

--
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"Gary White" <reply@newsgroup.please> wrote in message
news:r7l4435l2n0bnbdkmcpf1u2s5qk4dq7lju@4ax.com...
> On Wed, 9 May 2007 14:59:44 -0400, "Murray *ACE*"
> <forums@HAHAgreat-web-sights.com> wrote:
>
>>That's what I thought, too. But I solved the problem by moving the PHP to
>>the include file!
>
>
> Curious. That sounds like one of those solutions that shouldn't work,
> but the problem goes away when you do it.
>
> Gary

Report · May 10, 2007

On Wed, 9 May 2007 20:26:02 -0400, "Murray *ACE*"
<forums@HAHAgreat-web-sights.com> wrote:

>Yeah - of course, I changed everything about it, too!

I'm still bewildered by what caused the problem. If I have two files,
first (includevar.php) like this:

<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01//EN"
" http://www.w3.org/TR/html4/strict.dtd">
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1">
<title>Variable in an Include Test</title>
<meta http-equiv="imagetoolbar" content="no">
</head>
<body>
<?php
$x='This is $x defined in includevar.php';
include 'foo.php';

print "$y";
?>
</body>
</html>

And foo.php like this:

<?php
$y='This is $y defined in foo.php';
print "foo.php says $x \n";
?>

The output I get is:

foo.php says This is $x defined in includevar.php
This is $y defined in foo.php

Clearly, the variable defined in the parent file is recognized in
foo.php and the variable defined in foo.php is recognized in the parent
file. Hence my bewilderment.

Gary

Report · May 10, 2007

Mine, too. But when you don't know what you are doing, sometimes the 'proof
of concept' tests you design are a bit off-kilter, doncha know? I can only
guess that i was suffering from that.

I'll verify that I correctly reported this and post back....

--
Murray --- ICQ 71997575
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"Gary White" <reply@newsgroup.please> wrote in message
news:2p6643pe4k3ncefjuhf2l2v3huvanuqfgm@4ax.com...
> On Wed, 9 May 2007 20:26:02 -0400, "Murray *ACE*"
> <forums@HAHAgreat-web-sights.com> wrote:
>
>>Yeah - of course, I changed everything about it, too!
>
>
> I'm still bewildered by what caused the problem. If I have two files,
> first (includevar.php) like this:
>
> <!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01//EN"
> " http://www.w3.org/TR/html4/strict.dtd">
> <html>
> <head>
> <meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1">
> <title>Variable in an Include Test</title>
> <meta http-equiv="imagetoolbar" content="no">
> </head>
> <body>
> <?php
> $x='This is $x defined in includevar.php';
> include 'foo.php';
>
> print "$y";
> ?>
> </body>
> </html>
>
> And foo.php like this:
>
> <?php
> $y='This is $y defined in foo.php';
> print "foo.php says $x \n";
> ?>
>
> The output I get is:
>
> foo.php says This is $x defined in includevar.php
> This is $y defined in foo.php
>
> Clearly, the variable defined in the parent file is recognized in
> foo.php and the variable defined in foo.php is recognized in the parent
> file. Hence my bewilderment.
>
> Gary

Report · May 10, 2007

On Thu, 10 May 2007 11:38:48 -0400, "Murray *ACE*"
<forums@HAHAgreat-web-sights.com> wrote:

>I'll verify that I correctly reported this and post back....

The first thing I'd look for is whether the variable is defined within a
function, or the reference to it is inside a function. For example:

$var="Hello";
doit();

function doit(){
echo $var;
}

In this case, the variable is defined outside the function, but you're
trying to access it inside a function. The variable is out of scope
there. To fix this type of problem, you can do one of two things. First,
you could pass the variable as a parameter:

$var="Hello";
doit($var);

function doit($x){
echo $x;
}

The other would be to declare the variable as global in the function:

function doit(){
global $var;
echo $var;
}

The first method is preferable because it has the least chance of
disturbing other code that might use the variable.

Gary