How to execute a php script from another php script by using the shell?

Question

I have a php file called sample.php with the following content:

<?php
echo "Hello World!";
?>

And what I want to do, is to run this php script using a second php script.

I think shell_exec could help me, but I don't know its syntax.

By the way, I want to execute this files with cpanel. So I have to execute the shell.

Is there any way to do this?

Solution 1

If you need to write a php file's output into a variable use the ob_start and ob_get_contents functions. See below:

<?php
    ob_start();
    include('myfile.php');
    $myStr = ob_get_contents();
    ob_end_clean();
    echo '>>>>' . $myStr . '<<<<';
?>

So if your 'myfile.php' contains this:

<?php
    echo 'test';
?>

Then your output will be:

>>>>test<<<<

Solution 2

You can use cURL for remote requests. The below is from php.net:

<?php
// create a new cURL resource
$ch = curl_init();

// set URL and other appropriate options
curl_setopt($ch, CURLOPT_URL, "http://www.example.com/");
curl_setopt($ch, CURLOPT_HEADER, 0);

// grab URL and pass it to the browser
curl_exec($ch);

// close cURL resource, and free up system resources
curl_close($ch);
?>

Here's a good tutorial: http://www.sitepoint.com/using-curl-for-remote-requests/

Consider watching this YouTube video here as well: http://www.youtube.com/watch?v=M2HLGZJi0Hk

Solution 3

You can try this:

Main PHP file

<?php
// change path/to/php according to how your system is setup
// examples: /usr/bin/php or /opt/lampp/bin/php
echo shell_exec("/path/to/php /path/to/php_script/script.php");
echo "<br/>Awesome!!!"
?>

Secondary PHP file

<?php
echo "Hello World!";
?>

Output when running Main PHP file

Hello World!
Awesome!!!

Hope it helps you.

Solution 4

Try this:

<?php
// change path/to/php according to how your system is setup
// examples: /usr/bin/php or /opt/lampp/bin/php
$output = shell_exec("/path/to/php /path/to/php_script/script.php");
print_r($output);
?>

This May Help you.

Solution 5

It's important to stress that including/executing user-generated code is dangerous. Using a system call (exec, shell_exec, system) instead of include helps separate the execution context, but it's not much safer. Consider proper sanitation or sand-boxing.

With that in mind, here is a working example including generating the (temporary) file, executing it, and cleanup:

<?php
   // test content  
   $code = <<<PHP
   echo "test";
PHP;

   // create temporary file
   $d=rand();
   $myfile="$d.php";
   file_put_contents($myfile,"<?php\n$code\n?>");

   // start capture output
   ob_start();

   // include generate file
   // NOTE: user-provided code is unsafe, they could e.g. replace this file.
   include($myfile);

   // get capture output
   $result = ob_get_clean();

   // remove temporary file
   unlink($myfile);

   // output result
   echo "================\n" . $result . "\n================\n" ;

Output:

================
test
================

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Comments

[Sanjay Rathod]

It is also return empty data. How can i find what is the error

[Sanjay Rathod]

Hi i had tried your suggestion. It worked. But i am taking data from editor and storing that data into file using this code $d=rand(); $myfile=$d.".php"; //file_put_contents($myfile,"code: ",FILE_APPEND); file_put_contents($myfile,"<?php "."\n",FILE_APPEND); file_put_contents($myfile,$code."\n",FILE_APPEND); file_put_contents($myfile,"?>"."\n",FILE_APPEND); so you can understand from my code that each time the file name should be different. i had tried this include($myfile); but it returns an error that no such file is found. So how can i do this

[sagunms]

If "Hello World!" above is not displayed, the error be displayed instead because of the "echo" before "shell_exec".

[Randy]

I'm not sure I see the problem. Just include the correct file. If it's named differently every time shouldn't you have a handle to the name of it somewhere?

[raphael75]

He is trying to have this execute at runtime.