How to pass parameters from bash to php script?

Question

I have done a a bash script which run php script. It works fine without parameters but when I add parameters (id and url), there are some errors:

PHP Deprecated:  Comments starting with '#' are deprecated in /etc/php5/cli/conf                                                                                        .d/mcrypt.ini on line 1 in Unknown on line 0
Could not open input file: /var/www/dev/dbinsert/script/automatisation.php?                                                                                        id=1

I run php script from the bash like this:

php /var/www/dev/dbinsert/script/automatisation.php?id=19&url=http://bkjbezjnkelnkz.com

score 33 · Accepted Answer · edited Oct 04 '12 at 18:07

33

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.

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

edited Oct 04 '12 at 18:07

Zuul

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answered Jul 21 '11 at 16:47

Tino Didriksen

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How would you pass a directory into the php script? – Michael May 12 '16 at 19:20
1

@Michael, same way, just read it from the `$argv` array. Also, thread necromancy is bad. – Tino Didriksen May 14 '16 at 18:34
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I'm not sure if I did something wrong, or if the php has changed, but it is unnecessary to use the `--`. When I use `$argv[1]`, it just reads the `--`. – ZomoXYZ May 24 '16 at 00:40

Bastion · Answer 2 · 2017-08-06T23:29:08.963

-- Option 1: php-cgi --

Use 'php-cgi' in place of 'php' to run your script. This is the simplest way as you won't need to specially modify your php code to work with it:

php-cgi -f /my/script/file.php id=19 myvar=xyz

-- Option 2: if you have a web server --

If the php file is on a web server you can use 'wget' on the command line:

wget 'http://localhost/my/script/file.php?id=19&myvar=xyz'

OR:

wget -q -O - "http://localhost/my/script/file.php?id=19&myvar=xyz"

-- Accessing the variables in php --

In both option 1 & 2 you access these parameters like this:

$id = $_GET["id"];
$myvar = $_GET["myvar"];

score 6 · Answer 3 · answered Jul 21 '11 at 16:54

You can't pass GET query parameters to the PHP command line interface. Either pass the arguments as standard command line arguments and use the $argc and $argv globals to read them, or (if you must use GET/POST parameters) call the script through curl/wget and pass the parameters that way – assuming you have the script accessible through a local web server.

This is how you can pass arguments to be read by $argc and $argv (the -- indicates that all subsequent arguments should go to the script and not to the PHP interpreter binary):

php myfile.php -- argument1 argument2

You can pass GET query parameters if you use php-cgi :) See examples here: https://serverfault.com/questions/187025/how-to-pass-get-variables-to-a-php-script-via-the-command-line — Bastion, Apr 21 '17 at 00:07

How to pass parameters from bash to php script?

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