Shell Run/Execute PHP Script with Parameters

Shell run/execute php script with parameters

test.php:

<?php
print_r($argv);
?>

Shell:

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

Run PHP script from CLI and pass parameters

You need to run PHP:

php /my/script.php arg1 arg2

Then in your PHP file:

<?php
// It's 0 based, but arg1 will be at index 1!
var_dump($argv);

Pass a variable to a PHP script running from the command line

The ?type=daily argument (ending up in the $_GET array) is only valid for web-accessed pages.

You'll need to call it like php myfile.php daily and retrieve that argument from the $argv array (which would be $argv[1], since $argv[0] would be myfile.php).

If the page is used as a webpage as well, there are two options you could consider. Either accessing it with a shell script and Wget, and call that from cron:

#!/bin/sh
wget http://location.to/myfile.php?type=daily

Or check in the PHP file whether it's called from the command line or not:

if (defined('STDIN')) {
  $type = $argv[1];
} else {
  $type = $_GET['type'];
}

(Note: You'll probably need/want to check if $argv actually contains enough variables and such)

How to pass parameters from bash to php script?

Call it as:

php /path/to/script/script.php -- 'id=19&url=http://bkjbezjnkelnkz.com'

Also, modify your PHP script to use parse_str():

parse_str($argv[1]);

If the index $_SERVER['REMOTE_ADDR'] isn't set.

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More advanced handling may need getopt(), but parse_str() is a quick'n'dirty way to get it working.

How to call PHP file from a shell script file

In your php file named test.php, for example

<?php
//like programs in c language, use $argc, $argv access command line argument number and arguments, uncomment below two line to dump $argc and $argv
//var_dump($argc); //an integer
//var_dump($argv); //an array with arguments
//use args and do anything you want
echo "do my job\n";
exit(0);

then create a shell script named test.sh

#! `which bash`
php=`which php`
i=10
while [[ $i -ge 0 ]];
do  
$php test.php 1 2
((i--))
done

put the two files into the same directory. Then run command in the terminal

bash test.sh

Execute a php script (with arguments) within another php script

You are not grabbing the output from that command. That's why you see nothing although the command was executed. There are several ways how to do that. This are the most common:

// test.php
<?php
    $value = 123;
    // will output redirect directly to stdout
    passthru("php -f test1.php $value");

    // these functions return the outpout
    echo shell_exec("php -f test1.php $value");
    echo `php -f test1.php $value`;
    echo system("php -f test1.php $value");

    // get output and return var into variables
    exec("php -f test1.php $value", $output, $return);
    echo $output;
?>

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