use globals to pass the information to other pages..
http://www.php.net/manual/en/language.variables.php
Ben [EMAIL PROTECTED] wrote in message
news:[EMAIL PROTECTED]...
Ok, I'm sort of a beginner at this, so cut me some slack. :-) Ok, I have
an
if then statement that after you click a link
include (http://www.site.com/home.php;);
}
else{
include (http://www.site.com/; . $page . .php);
}
?
This code is in the index.php file. When home.php is included, I
want to echo a variable like $title on the index.php page. But
$title is defined on home.php. For some reason, I
Ben wrote:
Ok, I'm sort of a beginner at this, so cut me some slack. :-) Ok, I have an
if then statement that after you click a link it looks like this:
blah?page=pagename It looks something like this:
?php
if(!$page) {
include (http://www.site.com/home.php;);
}
else{
include
How would I do this then?
Leif K-Brooks [EMAIL PROTECTED] wrote in message
news:[EMAIL PROTECTED]
Ben wrote:
Ok, I'm sort of a beginner at this, so cut me some slack. :-) Ok, I have
an
if then statement that after you click a link it looks like this:
blah?page=pagename It looks something
Ben wrote:
How would I do this then?
include('foo.php');
--
The above message is encrypted with double rot13 encoding. Any unauthorized attempt
to decrypt it will be prosecuted to the full extent of the law.
--
PHP General Mailing List (http://www.php.net/)
To unsubscribe, visit:
thanks a bunch
Leif K-Brooks [EMAIL PROTECTED] wrote in message
news:[EMAIL PROTECTED]
Ben wrote:
How would I do this then?
include('foo.php');
--
The above message is encrypted with double rot13 encoding. Any
unauthorized attempt to decrypt it will be prosecuted to the full extent of
But how would I go about using the IF THEN statement but allowing the page
to see the variables.
Leif K-Brooks [EMAIL PROTECTED] wrote in message
news:[EMAIL PROTECTED]
Ben wrote:
How would I do this then?
include('foo.php');
--
The above message is encrypted with double rot13
Ben wrote:
But how would I go about using the IF THEN statement but allowing the page
to see the variables.
Not sure if I know what you mean, but:
if($foo == 'bar'){
include('foo.php');
}
--
The above message is encrypted with double rot13 encoding. Any unauthorized attempt
to decrypt it
?
function test1()
{$hello=hello;}
function test2()
{test1();
print $hello;}
test2();
?
This should print hello. But it does'nt work. Any idea?
Read here to find out why:
http://www.php.net/manual/en/language.variables.scope.php
---John Holmes...
--
PHP General Mailing List
?
function test1()
{return hello;}
function test2()
{$hello = test1();
print $hello;}
test2();
?
or make it (argh) global
-Original Message-
From: Michael P. Carel [mailto:[EMAIL PROTECTED]]
Sent: Tuesday, June 25, 2002 3:14 PM
To: php
Subject: [PHP] using variables in a function
thanks
?
function test1()
{return hello;}
function test2()
{$hello = test1();
print $hello;}
test2();
?
or make it (argh) global
-Original Message-
From: Michael P. Carel [mailto:[EMAIL PROTECTED]]
Sent: Tuesday, June 25, 2002 3:14 PM
To: php
Subject: [PHP] using
Hi
you need:
img src=?echo $ulo?
or
img src=?=$ulo?
Tom
At 05:39 PM 16/12/01, David Killingsworth wrote:
I'm trying to determing why this won't work.
I wish to assign graphics to variables then call them
within the HTML on my .php pages.
--test.php--
?
$ulo = /images/ulo.gif;
?
David,
Your code should read:
html
body
img src=? echo($ulo); ?
/body
/html
-Original Message-
From: David Killingsworth [mailto:[EMAIL PROTECTED]]
Sent: Sunday, December 16, 2001 2:39 AM
To: [EMAIL PROTECTED]
Subject: [PHP] using variables
I'm trying to determing why this won't
13 matches
Mail list logo
