Problem using file_get_contents()

habbardone

Hi,

I am trying to make my sites more uniform by using
some code from my central website.

I am trying to use file_get_contents() followed by eval();
working properly so I can have several websites use just one
bunch of pages.

The php file is on the webpage: buymobilewebsite.org

and the website ( or other directory ) which has the
common files is mobi6.net.

Both are on the same drive on the same physical server under
the /home directory.

I have done a simple version for testing:

The php:

http://buymobilewebsite.org/test.php

<?php 
/*
*  index.php 
*
*/
$site = "HELLO - SITE";

echo "Site: $site<br>";

$test1 = file_get_contents("/home/mobi463d/public_html/mobile_sites/test1.txt"); 
$the_test1 = eval($test1);
echo $the_test1;

$test2 = file_get_contents("/home/mobi463d/public_html/mobile_sites/test2.txt"); 
$the_test2 = eval($test2);
echo $the_test2;
?>

And the first text file:

test1.txt

<?php
/*
*  test1.txt
*
*/

echo "<br>Test1 - Site: $site</b>";
?>

And the first text file:

test2.txt

<?php
/*
*  test2.txt
*
*/

echo "<br>This is Test2 - Site : $site</b>";
?>

All I get is the output from my first echo.
The two text files do not seem to get read or echoed ?? 😕😕

See it here:

http://buymobilewebsite.org/test.php

EDIT:

I also tried using the url, but no change 🙁

<?php 
/*
*  index.php 
*
*/
$site = "HELLO - SITE";

echo "Site: $site<br>";

$test1 = file_get_contents("http://mobi6.net/mobile_sites/test1.txt"); 
$the_test1 = eval($test1);
echo $the_test1;

$test2 = file_get_contents("http://mobi6.net/mobile_sites/test2.txt"); 
$the_test2 = eval($test2);
echo $the_test2;

?>

http://buymobilewebsite.org/test.php

Any ideas why this isn't working ?

Thanks.

Derokorian

First thing I would say is why are you using eval? its EVIL.

php.net wrote:
The eval() language construct is very dangerous because it allows execution of arbitrary PHP code. Its use thus is discouraged. If you have carefully verified that there is no other option than to use this construct, pay special attention not to pass any user provided data into it without properly validating it beforehand.

Second when using eval:

php.net wrote:
The code mustn't be wrapped in opening and closing PHP tags, i.e. 'echo "Hi!";' must be passed instead of '<? echo "Hi!"; >'. It is still possible to leave and reenter PHP mode though using the appropriate PHP tags, e.g. 'echo "In PHP mode!"; ?>In HTML mode!<? echo "Back in PHP mode!";'.

TL😃R - don't use initial opening or final closing tags.

habbardone

Hi,

First thing I would say is why are you using eval? its EVIL.

If you can recommend another way to do this job,
I am all ears. 😕

BUT - the only files I am using are ones that I have written
so there is no security issue.

Thanks.

Derokorian

How about just making them php files and all you have to do is include them.
file1.php

<?php
/*
*  file1.php
*
*/

echo "<br>Test1 - Site: $site</b>";
?>

file2.php

<?php
/*
*  file2.php
*
*/

echo "<br>Test2 - Site: $site</b>";
?>

index.php

<?php 
/* 
*  index.php 
* 
*/ 
$site = "HELLO - SITE"; 

echo "Site: $site<br>"; 

include 'file1.php';
include 'file2.php';

?>

After all why put PHP code in a txt file and then eval it when you can just make a php file.

Also the way you are using it makes no sense, the eval of both text files would not have a return value, its only command is an echo statement so there is nothing stored in the variables to then echo after the eval.

habbardone

Good point,
and of course that's what I initially tried.

But apparently you can not access and process a php file that
is on a different part of the server.

Someone suggested that a way round this was to use
file_get_contents() followed by eval();.

So that's what I have been trying, without success.

I also tried this:

<?php 
/*
*  index.php 
*
*/
$site = "HELLO - SITE";

echo "Site: $site<br>";

eval(file_get_contents('http://mobi6.net/mobile_sites/test1.txt')); 

eval(file_get_contents('http://mobi6.net/mobile_sites/test2.txt')); 

?>