Redirect file output in a variable?

[TheJuice]

Hi,
I have a problem using PHPNuke:
If I want to print things to the screen in a block, I have to put everything in a variable called content:

$content = "I write this in the block.<br>";

but what I want to do is include a file in the block. In my file I print some things to the screen. But how could I do to display what my file outputs, in the block???

I'd like to do something like this:

$content = include('my_file.php');

I know this is not possible, so how could I redirect the output of my_file.php in the variable called $content?

Thanks.

[herve]

Hello,

well, you can change your code in my_file.php
in such a way to include everything within a
single function.

(new) content of my_file.php :

<?PHP

sub my_fct()
{
// instead of print everything with
print "Hello world";

// now you say :
$WhatIWant = "Hello world";

// and at the end :

return $WhatIWant

}

?>

that way, in your main program, you just wrote :

$content = my_fct();

Hervé.

[TheJuice]

Well thanks,
but I was asking, mainly because I didn't want to rewrite my code: I use templates, and so I don't use 'print' or 'echo' functions...

For now I rewrote my code to do as you said, but I still really want to to know if it's possible to do as I would like.

I know that under Unix, it is possible to redirect output in a file with '>': ls > list_of_my_files for example...

But in PHP, doing this kind of thing may not be possible ?

[herve]

Hi,

well, I quite sure you can do that with
system() or exec() call; but you PHP code
has to be a "real" PHP application, since
you're not executing it from a web server
but like a regular program (like a shell script,
or a *.exe prg under Windows)

so, if you are under Windows, something like
that :

exec ("/php/php.exe my_fct.php > myfile.txt");

involking the PHP interpreter first, redirecting
to a regular text file after.

and then :

$content = @file("myfile.txt");

to get back the output of the PHP script.

Hervé.

[TheJuice]

wow that could work.
But I don't know what a "real" PHP application is?

anyway, this could be a solution thanks a lot :-D

[herve]

by "real PHP program", I mean that normally
PHP program are launch by a web server like
Apache or IIS.

Which means that you have configured your web
server to "know" that to lauch a .php script,
it has to send it to the PHP interpreter.

It's not the web server itself which interpret
the PHP code.

In your context, that mean you have to
launch the PHP script "by hand", that's why
you need to lauch the PHP interpreter first,
telling him which PHP script you want to execute.

The PHP interpreter is an executable application
under your favorite OS.

Under Unix, you can also specify with a special
first line, who is responsible for interpreting
your file.

For instance , in a shell script, the first line
is often :

#!/bin/sh

which means that the command interpreter has
to lauch the "sh" program for interpreting the
current shell script.

You can do the same kind of things for PHP script
too.

Hervé.

[TheJuice]

Ok it's clearer now.
I don't know if I'll do that, but at least I know I could :-)

Maybe the simplest way is to change my code indeed...

Thanks for your help 😃