How to read an array from another php page

Question

I have the below mentioned code (actually, configuration) written in a file (config.php) and i want to get the content written in config.php in another file (check.php)

Codes written in config.php:

<?php
$CONFIG = array (
  'id' => 'asd5646asdas',
  'dbtype' => 'mysql',
  'version' => '5.0.12',
);

Code written in check.php to get the content is:

$config_file_path = $_SERVER['DOCUMENT_ROOT'] . 'config.php';
$config = file_get_contents($config_file_path);

Using the above code i am getting the output as a string and i wanted to convert it into an array. To do so, i have tried the below code.

$config = substr($config, 24, -5); // to remove the php starting tag and other un-neccesary things

$config = str_replace("'","",$config);
$config_array = explode("=>", $config);

Using the above code, i am getting an output like:

Array
(
    [0] =>   id 
    [1] =>  asd5646asdas,
  dbtype 
    [2] =>  mysql,
  version 
    [3] =>  5.0.12
)

which is incorrect.

Is there any way to convert it into an array. I have tried serialize() as well as mentioned in accessing array from another page in php , but did not succeed.

Any help on this will be appreciated.

You may include the file. If you want to parse the array, you might check this question. There is a custom php parser and a regex solution in the answers — HamZa
– HamZa, Commented Jun 21, 2013 at 12:31

HamZa · Accepted Answer · 2013-06-21 12:35:55Z

8

You don't need file_get_contents:

require_once 'config.php'
var_dump($CONFIG);

edited Jun 21, 2013 at 12:35

HamZa

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answered Jun 21, 2013 at 12:33

Shuro

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Comments

Karl Adler · Accepted Answer · 2013-06-21 12:38:49Z

Use include or require to include and execute the file. The difference is just what happens when the file doesn't exist. inlcude will throw a warning, require an error. To make sure just to load (and execute) the file the first time you can also use include_once, require_one php.net If you are somehow not able to include the file (what reasons ever) take a look at the eval() function to execute php code from string. But this is not recommendet at all because of security reasons!

require_once('config.php');
include_once('config.php');
require('config.php');
include('config.php');

Adam · Accepted Answer · 2013-06-21 12:29:44Z

3

I'm very confused at your approach, if you just do this:

include(config.php);

Then your $CONFIG variable will be available on other pages.

print_r($CONFIG);

Take a look on the PHP website which documents the include function.

answered Jun 21, 2013 at 12:29

Adam

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Comments

KAsh · Accepted Answer · 2013-06-21 12:33:09Z

1

if you are using codeigniter no need to include the config file, you can access using $this, but if you are using plain php, u need to use include() or require() or include_once() or require_once()

answered Jun 21, 2013 at 12:33

KAsh

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1 Comment

Debashis Over a year ago

Yes, i am using codeigniter but the config.php file is located outside of the framework. I will try your suggestion.

DevZer0 · Accepted Answer · 2013-06-21 12:29:09Z

0

simple

include_file "config.php";

answered Jun 21, 2013 at 12:29

DevZer0

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1 Comment

Debashis Over a year ago

Oh, yes... I really forget this approach.

hek2mgl · Accepted Answer · 2013-06-21 12:34:47Z

-1

How could you fail to read about the require_once directive? :)

$config = file_get_contents($config_file_path);

Must be:

require_once $config_file_path;
// now you can use variables, classes and function from this file
var_dump($config);

If you follow the link to the require_once manual page please read also about include and include_once. As a PHP developer you MUST know about this. It's elementary

edited Jun 21, 2013 at 12:34

answered Jun 21, 2013 at 12:29

hek2mgl

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