Running python script in Laravel

Question

So, I am trying to run a python script in my Laravel 5.3.

This function is inside my Controller. This simply passes data to my python script

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Answer 1

The solution that worked for me was to add 2>&1 to the end.

shell_exec("python path/to/script.py 2>&1");

The issue I was having was no error and no response, I had the wrong path to the script but no way of knowing. 2>&1 will redirect the debug info to be the result.

In the shell, what does " 2>&1 " mean?

Answer 2

Use Symfony Process. https://symfony.com/doc/current/components/process.html

Install:

composer require symfony/process

Code:

use Symfony\Component\Process\Process;
use Symfony\Component\Process\Exception\ProcessFailedException;

$process = new Process(['python', '/path/to/your_script.py']);
$process->run();

// executes after the command finishes
if (!$process->isSuccessful()) {
    throw new ProcessFailedException($process);
}

echo $process->getOutput();

Answer 3

Would be handy to use Symfony Process. https://symfony.com/doc/current/components/process.html

Make sure symphony is available to use in your project

composer show symphony/process

If not installed do composer require symfony/process

And then do some thing like

use Symfony\Component\Process\Process;
use Symfony\Component\Process\Exception\ProcessFailedException;

//$process = new Process('python /path/to/your_script.py'); //This won't be handy when going to pass argument
$process = new Process(['python','/path/to/your_script.py',$arg(optional)]);
$process->run();

// executes after the command finishes
if (!$process->isSuccessful()) {
    throw new ProcessFailedException($process);
}

echo $process->getOutput();