How to use parameters in commanline PHP

[lhg1]

Hi

I have a PHP script that I want to use from the commandline, and it has a parameter.

Usally I use this in a browser

http://host.dk/script.php?tid=2

where I in the script has this.

if(!isset($_GET['tid'])) {
        $tid = "1";
} else {
        $tid = $_GET['tid'];
}

I can run it on the command line without the parametes like this.
php -c /etc/php5/apache2/ ./script.php

But can anybody tell me how to get the parameter "tid" on the one from the commandline?

Regards
Lars

 

[Noway2]

PHP uses the argc and argv arrays like C when operated from the command line. See this link for an example. Specifically look at the section titled: diego dot rodrigues at poli dot usp dot br 02-May-2005 08:29

 

[feherke]

Hi

Either process the [tt]$argv[/tt] array yourself or pass it through [tt]getopt()[/tt] :

#!/usr/bin/php
[teal]<?php[/teal]

[b]echo[/b] [green][i]'Raw parameters : '[/i][/green][teal];[/teal]
[COLOR=darkgoldenrod]print_r[/color][teal]([/teal][navy]$argv[/navy][teal]);[/teal]

[navy]$opt[/navy][teal]=[/teal][COLOR=darkgoldenrod]getopt[/color][teal]([/teal][b]null[/b][teal],[/teal][b]array[/b][teal]([/teal][green][i]'tid:'[/i][/green][teal]));[/teal]

[b]echo[/b] [green][i]'Processed parameters : '[/i][/green][teal];[/teal]
[COLOR=darkgoldenrod]print_r[/color][teal]([/teal][navy]$opt[/navy][teal]);[/teal]

[blue]master #[/blue] ./lhg1.php --tid=foo bar
Raw parameters : Array
(
    [0] => ./lhg1.php
    [1] => --tid=foo
    [2] => bar
)
Processed parameters : Array
(
    [tid] => foo
)

Feherke.

http://feherke.github.com/