include-problem

Hi!

I'm having a problem with including files. What I want to achieve is to
execute a PHP-script on another server, and then to include the result
(which will be XML-output) in another PHP-script (currently on my local
computer).

On the server I have the file http://server.com/test/echo.php with the
content

---
<?php
echo 'xyz';
?>
---

Locally I've a file with the following content:

---
<?php
echo "!!!";
include ("http://server.com/test/echo.php");
echo "???";
?>
---

I was expecting the output from my locally testfile to be something
like:

---
!!!???
---

but rather it is

---
!!!xyz???
---

I've also tried with a $fp = readfile("http…") with the same result,
which is output of the echo-statement in the remote file which I am
expecting to be evaluated remotely.

How can I do to include the PHP-script and have it to be ran before it
is included?

Sincerely

Victor

Victor Spång Arthursson wrote:

> Hi!
>
> I'm having a problem with including files. What I want to achieve is
> to execute a PHP-script on another server, and then to include the
> result (which will be XML-output) in another PHP-script (currently on
> my local computer).
>
> On the server I have the file http://server.com/test/echo.php with the
> content
>
> ---
> <?php
> echo 'xyz';
> ?>
> ---
>
> Locally I've a file with the following content:
>
> ---
> <?php
> echo "!!!";
> include ("http://server.com/test/echo.php");
> echo "???";
> ?>
> ---
>
> I was expecting the output from my locally testfile to be something like:
>
> ---
> !!!???
> ---
>
> but rather it is
>
> ---
> !!!xyz???
> ---
>
> I've also tried with a $fp = readfile("http…") with the same result,
> which is output of the echo-statement in the remote file which I am
> expecting to be evaluated remotely.
>
> How can I do to include the PHP-script and have it to be ran before it
> is included?
>
> Sincerely
>
> Victor


This is a perfectly normal behaviour ! See www.php.net/include to
understand what this function does. (comparing to www.php.net/require)

Hope this helps,

--
Cordialement,
---------------------------
Sophie Mattoug
Développement web dynamique
sophie@mattoug.net
---------------------------

On Mon, 1 Dec 2003, Sophie Mattoug wrote:
> Victor Spång Arthursson wrote:
>
> > Hi!
> >
> > I'm having a problem with including files. What I want to achieve is
> > to execute a PHP-script on another server, and then to include the
> > result (which will be XML-output) in another PHP-script (currently on
> > my local computer).
> >
> > On the server I have the file http://server.com/test/echo.php with the
> > content
> >
> > ---
> > <?php
> > echo 'xyz';
> > ?>
> > ---
> >
> > Locally I've a file with the following content:
> >
> > ---
> > <?php
> > echo "!!!";
> > include ("http://server.com/test/echo.php");
> > echo "???";
> > ?>
> > ---
> >
> > I was expecting the output from my locally testfile to be something like:
> >
> > ---
> > !!!???
> > ---
> >
> > but rather it is
> >
> > ---
> > !!!xyz???
> > ---
> >
> > I've also tried with a $fp = readfile("http…") with the same result,
> > which is output of the echo-statement in the remote file which I am
> > expecting to be evaluated remotely.
> >
> > How can I do to include the PHP-script and have it to be ran before it
> > is included?
> >
> > Sincerely
> >
> > Victor
>
>
> This is a perfectly normal behaviour ! See www.php.net/include to
> understand what this function does. (comparing to www.php.net/require)

It's perfectly normal, yes, but it has nothing to do with include vs.
require.

I guess I don't really understand the question. I assume you realize that
an include 'http://server.com/file.php' is going to send an HTTP request
to server.com asking for file.php and if server.com is configured to
execute php for file.php then what will come back across the wire is the
result of php running the script in file.php. As such, when you do:

echo '!!!';
include 'http://server.com/file.php';
echo '???';

you will of course see: !!!xyz??? because that is exactly what you have
asked it to do. Print !!!, then send an HTTP request to server.com and
include the output of that script right here, and finally print out ???.
So I don't understand why this output is surprising you and I don't
understand your question about expecting it to be evaluated remotely.
file.php was of course evaluated remotely on server.com. If file.php had
written something to the filesysts, for example, then that something would
be on server.com not on your server.

-Rasmus

Rasmus Lerdorf wrote:
> On Mon, 1 Dec 2003, Sophie Mattoug wrote:
>> Victor Spång Arthursson wrote:
>>
>>> Hi!
>>>
>>> I'm having a problem with including files. What I want to achieve is
>>> to execute a PHP-script on another server, and then to include the
>>> result (which will be XML-output) in another PHP-script (currently
>>> on my local computer).
>>>
>>> On the server I have the file
> http://server.com/test/echo.php with
>>> the content
>>>
>>> ---
>>> <?php
>>> echo 'xyz';
>>>>
>>> ---
>>>
>>> Locally I've a file with the following content:
>>>
>>> ---
>>> <?php
>>> echo "!!!";
>>> include ("http://server.com/test/echo.php");
>>> echo "???";
>>>>
>>> ---
>>>
>>> I was expecting the output from my locally testfile to be something
>>> like:
>>>
>>> ---
>>> !!!???
>>> ---
>>>
>>> but rather it is
>>>
>>> ---
>>> !!!xyz???
>>> ---
>>>
>>> I've also tried with a $fp = readfile("http…") with the same result,
>>> which is output of the echo-statement in the remote file which I am
>>> expecting to be evaluated remotely.
>>>
>>> How can I do to include the PHP-script and have it to be ran before
>>> it is included?
>>>
>>> Sincerely
>>>
>>> Victor
>>
>>
>> This is a perfectly normal behaviour ! See www.php.net/include to
>> understand what this function does. (comparing to
>> www.php.net/require)
>
> It's perfectly normal, yes, but it has nothing to do with include vs.
> require.
>
> I guess I don't really understand the question. I assume you
> realize that an include 'http://server.com/file.php' is going
> to send an HTTP request to server.com asking for file.php and
> if server.com is configured to execute php for file.php then
> what will come back across the wire is the result of php
> running the script in file.php. As such, when you do:
>
> echo '!!!';
> include 'http://server.com/file.php';
> echo '???';
>
> you will of course see: !!!xyz??? because that is exactly
> what you have asked it to do. Print !!!, then send an HTTP
> request to server.com and include the output of that script
> right here, and finally print out ???.
> So I don't understand why this output is surprising you and I
> don't understand your question about expecting it to be
> evaluated remotely.
> file.php was of course evaluated remotely on server.com. If
> file.php had written something to the filesysts, for example,
> then that something would be on server.com not on your server.
>
> -Rasmus

You can use the output buffer functions to catch the "xyz" into a var:

<?php
print "!!!";
ob_start();
include 'http://server.com/test/echo.php';
$XML = ob_get_clean(); // or use ob_get_contents(); and ob_end_clean() for
PHP < 4.3
print "???";

print '[Between this you'll get your XYZ]';
print $XML;
print '[Between this you'll get your XYZ]';
?>

Hope it helps ya,
Wouter

On Mon, 1 Dec 2003, Wouter van Vliet wrote:
> <?php
> print "!!!";
> ob_start();
> include 'http://server.com/test/echo.php';
> $XML = ob_get_clean(); // or use ob_get_contents(); and ob_end_clean() for
> PHP < 4.3
> print "???";
>
> print '[Between this you'll get your XYZ]';
> print $XML;
> print '[Between this you'll get your XYZ]';
> ?>

Or just use file_get_contents() which would be more efficient than using
output buffering for this.

-Rasmus

On maandag 1 december 2003 15:23 Rasmus Lerdorf told the butterflies:
> On Mon, 1 Dec 2003, Wouter van Vliet wrote:
> > <?php
> > print "!!!";
> > ob_start();
> > include 'http://server.com/test/echo.php';
> > $XML = ob_get_clean(); // or use ob_get_contents(); and
> > ob_end_clean() for PHP < 4.3 print "???";
> >
> > print '[Between this you'll get your XYZ]'; print $XML; print
> > '[Between this you'll get your XYZ]'; ?>
>
> Or just use file_get_contents() which would be more efficient
> than using output buffering for this.
>
> -Rasmus

yes, probably. But isn't file_get_contents(); only implemented from php4.3
... might wanna try

join('', file());

or

$fd = fopen($FileName, 'r');
fread($fd, filesize($fd));

if you're running an older version.

-me.