test.php:

<?php
print_r($argv);
?>

Shell:

$ php -q test.php foo bar
Array
(
    [0] => test.php
    [1] => foo
    [2] => bar
)

Script:

<?php

// number of arguments passed to the script
var_dump($argc);

// the arguments as an array. first argument is always the script name
var_dump($argv);

Command:

$ php -f test.php foo bar baz
int(4)
array(4) {
  [0]=>
  string(8) "test.php"
  [1]=>
  string(3) "foo"
  [2]=>
  string(3) "bar"
  [3]=>
  string(3) "baz"
}

Also, take a look at using PHP from the command line.

Passing parameters to a shell script from PHP.

Its all about the "strings", and when to "double quote" for expansion.

<?php

/* exec("/csvexport.sh $table"); */

/* double quote here because you want PHP to expand $table */
/* Escape double quotes so they are passed to the shell because you do not wnat the shell to choke on spaces */
$command_with_parameters = "/path/csvexport.sh \"${table}\"";
$output_from_command = "";
$command_success = "";

/* double quote here because you want PHP to expand $command_with_parameters, a string */
exec("${command_with_parameters}", $output_from_command, $command_success);

/* or to keep it simple */
exec("/path/csvexport.sh \"${table}\"");


/* show me what you got */
echo"${command_success}\n${output_from_command}\n";

?>

BTW: I did not test this snippet.

Your call is failing because you're using a web-style syntax (?parameter=value) with a command-line invokation. I understand what you're thinking, but it simply doesn't work.

You'll want to use $argv instead. See the PHP manual.

To see this in action, write this one-liner to a file:

<?php print_r($argv); ?>

Then invoke it from the command-line with arguments:

php -f /path/to/the/file.php firstparam secondparam

You'll see that $argv contains the name of the script itself as element zero, followed by whatever other parameters you passed in.

You're starting your command with . - that will be interpreted as shell-source command, which is not what you want, obviously. Instead specify full path and do like:

$result = shell_exec('/bin/bash /full/path/to/script.sh param1 param2');
//var_dump($result);

-also, make sure your php user have permission to execute your sh-script and that PHP can use functions like exec() (they could be disabled by configuration)

Can you try something like the below code snippet.

private function runCom(){
    $command = 'nohup sudo python /home/pi/Python/RGB_Controller/write_color.py' .$HEX.'& > /dev/null 2>&1 & echo $!';
    exec($command ,$op);
    $this->pid = (int)$op[0];
}

Its like you build your command with necessary arguments as a string.

Then exec($command ,$op). Please check and let me know.

Which PHP binary are you using? The CLI or CGI? I suspect you need a CGI version of the binary for PHP to properly handle accept the environment variables and POST data if you pass that.

The php-cgi binary implements the CGI interface, which allows you to pass parameters on the command line:

php-cgi -f index.php left=1058 right=1067 class=A language=English

Which end up in $_GET:

Array
(
    [left] => 1058
    [right] => 1067
    [class] => A
    [language] => English
)

You may want to read up on how CGI works so you can implement that in your web server.

Ref: RFC3875

If you want to execute this script from PHP as well as from Java, then I'd take this approach:

• put your code which does the work into a function within the file "myfunc.php" (give it a more meaningful name). Make the function take whatever arguments it needs, but it shouldn't try to accss $_GET parameters because these will only exist when called through the web.
• now create a PHP file which includes "myfunc.php" and calls the function. This file can access $_GET and other web-specific variables, passing them as parameters to your function.
• To also be able to execute from Java, create a 3rd file called "mycommand.php", which you can execute from the command line. If you need to pass command line arguments, see documentation on $argv. This script should include "myfunc.php" and call the function, passing any necessary parameters.

Develop and test this last script by running your script from the command line:

$ php5 mycommand.php <args go here>

After you get it working like this, then you can invoke it from java by using this solution.

There is no reason to invoke a PHP command as a subprocess from a PHP script - using "include" as I've described above is easier and more efficient. However here is an example for testing purposes:

test.php (invoke this through the browser):

<?
exec("php5 test2.php", $ret);
foreach ($ret as $line) {
  print $line . '<br>';
}
?>

This script invokes a second PHP script called "test2.php" (in the same directory), and prints the output from that command.

test2.php simply produces some output:

<?  
print "foo\n"; 
print "bar\n"; 
print "glorp\n"; 
?>

You can use that code:

start_date="2014-10-05"
days=2
for i in $(seq 0 $days);
do
  php update_cron.php $(date "+%Y-%m-%d" --date="$start_date +$i days")
done

Explanation:

• start_date="2014-10-05": Sets the date where you start
• days=2: sets the number of days you want (note: 2 means 3 days in total, day 0 also counts)
• for i in $(seq 0 $days);: loops trough $days times
• php update_cron.php $(...): executes php update_cron.php with the argument
• $(date "+%Y-%m-%d" --date="$start_date +$i days"): formats the date and estimates the date from $start_date plus $i days

Presumably you're passing the arguments in on the command line as follows:

php /path/to/wwwpublic/path/to/script.php arg1 arg2

... and then accessing them in the script thusly:

<?php
// $argv[0] is '/path/to/wwwpublic/path/to/script.php'
$argument1 = $argv[1];
$argument2 = $argv[2];
?>

What you need to be doing when passing arguments through HTTP (accessing the script over the web) is using the query string and access them through the $_GET superglobal:

Go to http://yourdomain.example/path/to/script.php?argument1=arg1&argument2=arg2

... and access:

<?php
$argument1 = $_GET['argument1'];
$argument2 = $_GET['argument2'];
?>

If you want the script to run regardless of where you call it from (command line or from the browser) you'll want something like the following:

as pointed out by Cthulhu in the comments, the most direct way to test which environment you're executing in is to use the PHP_SAPI constant. I've updated the code accordingly:

<?php
if (PHP_SAPI === 'cli') {
    $argument1 = $argv[1];
    $argument2 = $argv[2];
}
else {
    $argument1 = $_GET['argument1'];
    $argument2 = $_GET['argument2'];
}
?>

Send $argv of serv.php from serv.php to update.php (see http://php.net/manual/en/reserved.variables.argv.php). I.e. in update.php you will have another $argv, so you have to send list of command line parameters in array with different name.

When calling a PHP script from the command line you can use $argc to find out how many parameters are passed and $argv to access them. For example running the following script:

<?php
    var_dump($argc); //number of arguments passed 
    var_dump($argv); //the arguments passed
?>

Like this:-

php script.php arg1 arg2 arg3

Will give the following output

int(4)
array(4) {
  [0]=>
  string(21) "d:\Scripts\script.php"
  [1]=>
  string(4) "arg1"
  [2]=>
  string(4) "arg2"
  [3]=>
  string(4) "arg3"
}

See $argv and $argc for further details.

To do what you want, lets say

php script.php arg1=4

You would need to explode the argument on the equals sign:-

list($key, $val) = explode('=', $argv[1]);
var_dump(array($key=>$val));

That way you can have whatever you want in front of the equals sign without having to parse it, just check the key=>value pairs are correct. However, that is all a bit of a waste, just instruct the user on the correct order to pass the arguments.

In case you don't want to modify running script, you can specify parameters using in -B parameter to specify code to run before the input file. But in this case you must also add -F tag to specify your input file:

php -B "\$_REQUEST = array('param1' => 'val1', 'param2' => 'val2');" -F yourscript.php