PHP - command line arguments in Windows - php

Question

I'm trying to run PHP from the command line under Windows XP.
That works, except for the fact that I am not able to provide parameters to my PHP script.
My test case:
echo "param = " . $param . "\n";
var_dump($argv);
I want to call this as:
php.exe -f test.php -- param=test
But I never get the script to accept my parameter.
The result I get from the above script:
PHP Notice: Undefined variable: param in C:\test.php on line 2
param = ''
array(2) {
[0]=> string(8) "test.php"
[1]=> string(10) "param=test"
}
I am trying this using PHP 5.2.6. Is this a bug in PHP 5?
The parameter passing is handled in the online help:
Note: If you need to pass arguments to your scripts you need to pass -- as the first argument when using the -f switch.
This seemed to be working under PHP 4, but not under PHP 5.
Under PHP 4 I could use the same script that could run on the server without alteration on the command line. This is handy for local debugging, for example, saving the output in a file, to be studied.

Answer 1

Why do you have any expectation that param will be set to the value?
You're responsible for parsing the command line in the fashion you desire, from the $argv array.

Answer 2

You can use the getopt() function.
Check blog post PHP CLI script and Command line arguments.

Answer 3

The parameter passing is handled in the online help Note: If you need to pass arguments to your scripts you need to pass -- as the first argument when using the -f switch. This seemed to be working under PHP 4, but not under PHP 5.
But PHP still doesn't parse those arguments. It just passes them to the script in the $argv array.
The only reason for the -- is so that PHP can tell which arguments are meant for the PHP executable and which arguments are meant for your script.
That lets you do things like this:
php -e -n -f myScript.php -- -f -n -e
(The -f, -n, and -e options after the -- are passed to file myScript.php. The ones before are passed to PHP itself).

Answer 4

If you want to pass the parameters similar to GET variables, then you can use the parse_str() function. Something similar to this:
<?php
parse_str($argv[1]);
?>
Would produce a variable, $test, with a value of <myValue>.

Answer 5

PHP does not parameterize your command line parameters for you. See the output where your second entry in ARGV is "param=test".
You most likely want to use the PEAR package Console_CommandLine: "A full featured command line options and arguments parser".
Or you can be masochistic and add code to go through your ARGV and set the parameters yourself. Here's a very simplistic snippet to get you started (this won't work if the first part isn't a valid variable name or there is more than 1 '=' in an ARGV part:
foreach($argv as $v) {
if(false !== strpos($v, '=')) {
$parts = explode('=', $v);
${$parts[0]} = $parts[1];
}
}

Answer 6

Command-line example:
php myserver.php host=192.168.1.4 port=9000
In file myserver.php, use the following lines:
<?php
parse_str(implode('&', array_slice($argv, 1)), $_GET);
// Read arguments
if (array_key_exists('host', $_GET))
{
$host = $_GET['host'];
}
if (array_key_exists('port', $_GET))
{
$port = $_GET['port'];
}
?>

Answer 7

$argv is an array containing all your commandline parameters... You need to parse that array and set $param yourself.
$tmp = $argv[1]; // $tmp="param=test"
$tmp = explode("=", $tmp); // $tmp=Array( 0 => param, 1 => test)
$param = $tmp[1]; // $param = "test";

Answer 8

You can do something like:
if($argc > 1){
if($argv[1] == 'param=test'){
$param = 'test';
}
}
Of course, you can get much more complicated than that as needed.

Answer 9

If you like living on the cutting edge, PHP 5.3 has the getOpt() command which will take care of all this messy business for you. Somewhat.

Answer 10

You could use something like
if (isset($argv[1]) {
$arg1 = $argv[1];
$arg1 = explode("=", $arg1);
$param = $arg1[1];
}
(How to handle the lack of parameters is up to you.)
Or if you need a more complex scenario, look into a command-line parser library, such as the one from Pear.
Using the ${$parts[0]} = $parts[1]; posted in another solution lets you override any variable in your code, which doesn’t really sound safe.

Answer 11

You can use the $argv array. Like this:
<?php
echo $argv[1];
?>
Remember that the first member of the $argv array (which is $argv[0]) is the name of the script itself, so in order to use the parameters for the application, you should start using members of the $argv[] from the '1'th index.
When calling the application, use this syntax:
php myscript.php -- myValue
There isn't any need to put a name for the parameter. As you saw, what you called the var_dump() on the $argv[], the second member (which is the first parameter) was the string PARAM=TEST. Right? So there isn't any need to put a name for the param. Just enter the param value.