PHP execute external script without waiting, while passing variables

Question

I'm making a function for my users where they can upload large XML files to synchronize with my database.

When a user uploads a file to upload.php, I want to start processing the data in the background with process.php, preferably from a shell command, and redirect the user to status.php, which shows the process of the synchronization.

I need to pass some variables to the process.php script while executing it, either at least one variable with the user id and put the other variables into a text file, (Would probably prefer this so I wont have to put to much data into the exec() command.) or the user id and a bunch of $_POST variables.

One solution I had in mind is executing the PHP script like this:

exec("php -f ./process.php > /dev/null 2>/dev/null &");

This allows me to lock away process.php from http access, which is good since it's a process taking script. The only thing I need here is to pass a variable somehow, but i don't know how to do it.

So my main question is:

How do i pass a variable in the above solution?

Or do any of you have a better solution to doing this? Possibly one where i wont have to go through exec()? Keep in mind that i do not want the user to wait for the script to execute, and i need to pass at least one variable.

Update: For future reference, remember to use escapeshellarg() when passing arguments through exec() or likewise functions.

jbalde · Accepted Answer · 2011-07-16 22:27:25Z

5

You test use it

exec("php -f ./process.php var1 var2 > /dev/null 2>/dev/null &");

And if you like get these variables values can acces with global variable $argv. If you print this var show same:

print_r($argv);


Array
(
    [0] => process.php
    [1] => var1
    [2] => var2
)

answered Jul 16, 2011 at 22:27

jbalde

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Comments

Sascha Galley · Accepted Answer · 2011-07-16 22:28:27Z

2

You can pass parameters like the following.

// call process.php
exec("php -f ./process.php foo=bar bar=foo > /dev/null 2>/dev/null &"); 

// process.php
if ($argc > 0) {
    for ($i=1;$i < $argc;$i++) {
        parse_str($argv[$i],$tmp);
        $_REQUEST = array_merge($_REQUEST, $tmp);
    }
}

var_dump($_REQUEST);

answered Jul 16, 2011 at 22:28

Sascha Galley

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Comments

Community · Accepted Answer · 2017-05-23 12:19:50Z

1

I don't really understood your goal, but to pass an argument to an PHP-script works similar to any other shell scripts. See: http://www.php.net/manual/en/features.commandline.usage.php (Example #2)

"When a user uploads a file […], I want to start processing the data in the background" - You can't access an upload before it is finished, in PHP using CGI.

edited May 23, 2017 at 12:19

CommunityBot

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answered Jul 16, 2011 at 22:24

feeela

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1 Comment

Kristoffer la Cour Over a year ago

The file will be saved before process.php is executed, don't worry. And thanks for the link.

Julian F. Weinert · Accepted Answer · 2012-08-14 14:39:31Z

Here is my solution.

The advantage is that you can use this script for command line usage as well as for regular web usage. In case you call it from cmd the $argv variable is set, so the parse_str() part extracts the variables and puts them into the $_GET array. In case you call it from the web the $argv is not set so the values come from the url.

// executingScript.php
// You have to make percent escaping yourself
exec("php -f ./executedScript.php foo=bar%20foo bar=foo%20bar > /dev/null 2>/dev/null &");

// executedScript.php
// The if-statement avoids crashing when calling the script from the web
if (isset($argv)) {
    parse_str(implode('&', array_slice($argv, 1)), $_GET);
}

This will make you able to access the variables as usual:

echo $_GET["foo"] // outputs "bar foo"
echo $_GET["bar"] // outputs "foo bar"

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