Is it possible to get returned data from a file in PHP?

Question

I am creating a reusable app, which requires some configurations stored in config.php. This configuration file does not have any function, but the configuration are returned as an array, eg

<?php

return [
    'db_name' => 'db_name',
    'password' => 'password',
     ........
];

I was hoping to create a function which includes this config file, and get each config set. An example is laravel config() function which returns a configuration value. Is there any way to acheive this?

I have tried

$filename = 'config.php';
include $filename;
$contents[] = file_get_contents($filename);

but this just gets whatever is in the file as a string.

Welcome to SO. Hint: please try searching before posting a new question. I tried super vague terms like "*php get common config*" and found many answers here already. — Don't Panic, Nov 24 '22 at 22:29
You do not want `file_get_contents()`. That will get the contents of a file. The methods described in the answers given here and the approach that uses `return` at https://stackoverflow.com/questions/14752470/creating-a-config-file-in-php bring the code in *config.php* into your script as if it was actually code written in your script. — user3425506, Nov 24 '22 at 23:14
You was so near, just needed `$config = include('config.php');` — José Carlos PHP, Nov 25 '22 at 00:44

score 1 · Answer 1 · answered Nov 24 '22 at 22:47

In a nutshell, this is possible if you assign the config array to a variable instead of return:

<?php

$config = [
    'db_name' => 'db_name',
    'password' => 'password',
     ........
];

Then, you can access the $config everywhere this file is included. You don't need file_get_contents in this case.

But please note that this solution adds a potential vulnerability to your application, literally allowing to execute an arbitrary code provided inside config.php. Instead of storing the configuration in a php file, consider using the dotenv component, or if this too complex for your app, use a harmless format such as JSON or YAML:

config.json:

{
    "db_name": "some_name",
    "password": "some_password",
    ...
}

then, in your code:

$filename = 'config.json';
$parsed = json_decode(file_get_contents($filename), true);
echo $parsed['db_name'];  // -> some_name
echo $parsed['password']; // -> some_password

If you write the code inside *config.php* yourself, it will be as safe or dangerous as any other code you write in your application. If you are importing it from somewhere else then I guess JSON or YAML are safer formats as there is a limit to what they can cause your program to do. — user3425506, Nov 24 '22 at 22:52
@user3425506 thanks for your note. I said _potential_ vulnerability meaning some day that config is supposed to be imported from some external source. — igops, Nov 24 '22 at 23:55
Using `return` is welcome to do this: `$config = include('config.php');` — José Carlos PHP, Nov 25 '22 at 00:43

user3425506 · Answer 2 · 2022-11-24T22:57:20.023

Edit I have just followed the link from where your question was closed (Creating a config file in PHP) and was interested to find out that the approach you have taken (using return) does appear to be valid and is recommended there. So, don't let me put you off that. I always do it like I explain below:

In config.php get rid of return and assign the array to a variable.

<?php
$config =  [
    'db_name' => 'db_name',
    'password' => 'password',
     ........
];

Then use require_once './config.php'; to get it into any .php script you like. You can then use $config in that script.

Note the path to config.php must be correct. I have assumed above that it is in the same directory as the file you are importing it into with require_once so the path used was ./config.php. If config.php was in the parent directory of the file using it then the path would have had two dots: ../config.php.

Is it possible to get returned data from a file in PHP?

2 Answers2