Pass PHP variables to a Bash script and then launch it

Question

So, basically, what I want to do is execute this Bash script :

#!/bin/bash

path=\"./\"
filename=\"Ellly.blend\"
file=${path}${filename}
description=\"Test o         


        

Answer 1

Note: I didn't do all the settings, just enough I hope you get the idea.

Option 1: Passing parameters

PHP:

$file = escapeshellarg($file);
$filename = escapeshellarg($filename);
// escape the others
$output = exec("./bashscript $file $filename $tags $private $password");

Bash:

#!/bin/bash
filename=$1
file=$2
description="Test of the api with a simple model"
token_api="ff00ff"
title="Uber Glasses"
tags=$3
private=$4
password=$5

...

---

Option 2: Using environment variables

PHP:

putenv("FILENAME=$filename");
putenv("FILE=$file");
putenv("TAGS=$tags");
putenv("PRIVATE=$private");
putenv("PASSWORD=$PASSWORD");

$output = exec('./bash_script');

Bash:

filename=$FILENAME
file=$FILE
description="Test of the api with a simple model"
token_api="ff00ff"
title="Uber Glasses"
tags=$TAGS
private=$PRIVATE
password=$PASSWORD

...