php CLI arguments Issues

Question

I'm working on a command-line script for a php app.

I can't get it to successfully accept params.

(via ssh)
If I type:
/: php testFile.php arg1 arg2 arg3

I receive the following errors: Notice: Undefined variable: agrc in /testFile.php on line 3 Notice: Undefined variable: agrv in /testFile.php on line 3

Here's the current code:

<?php
echo ini_get('register_argc_argv');
print_r( $agrc );print_r( $agrv );
?>

I also confirmed that the 'register_argc_argv' directive is TRUE

What am I over looking ?

Thanks in advance.

You check for the argc and argv setting, and then try to debug agrc and agrv. Look at those variable names again. — jszobody
– jszobody, Commented Jan 8, 2014 at 21:08

Pedro Lobito · Accepted Answer · 2014-01-08 22:00:57Z

2

Use $_SERVER['argv'] instead of only $agrv.

edited Jan 8, 2014 at 22:00

answered Jan 8, 2014 at 21:15

Pedro Lobito

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5 Comments

user2977468 Over a year ago

We don't use CGI anywhere in the app now, and the php-cgi binary is not on our server. I believe this can be done using the cli, as well as cgi. Have I misunderstood the cli ?

Pedro Lobito Over a year ago

Have tried to add #!/usr/bin/php -q to the first line of your file , even before the php tag (<?php) ? I've updated the answer, check it out.

user2977468 Over a year ago

Copy & paste your example and it fails - while . Also added the declaration line up top, to no avail. Possibly a directives issue ?

Pedro Lobito Over a year ago

Try your original code but use $_SERVER['argc'] instead of only $agrc. Does it work ?

user2977468 Over a year ago

A quick var_dmup shows me it's in $_SERVER['agrv'] , that'll work. Thanks Tuga !

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