How to pass command line arguments in PHP?

Question

Guys I’d like to do those scripts that pass arguments.

Ex. php script.php -email [email protected]

Where can I find documentation about this?

PS: I just want to know how to pass these arguments

Answer 1

Use $argv, thus:

var_dump($argv);

Is similar to c

Note that $argv[0] always returns the name of the script called, for example in Windows:

C:\Users\Guilherme\Desktop>php teste.php foo bar
Array
(
    [0] => C:\Users\Guilherme\Desktop\teste.php
    [1] => foo
    [2] => bar
)

To remove the first item you can use array_shift, thus:

<?php

$argumentos = $argv;
$script = array_shift($argumentos);

echo $script, '<br>';

echo 'Argumentos:<br>';

var_dump($argumentos);

Assuming you want to take specific elements you won’t even need the array_shift, if you know you want to take the first parameter and the second for example, just check with an IF() the amount (use the $argc to obtain the amount of parameters passed) and then take the specific index:

if ($argc >= 3) {
    $primeiro = $argv[1];
    $segundo = $argv[2];
} else {
    die('Parâmetros incorretos');
}

Documentation:

• http://php.net/manual/en/reserved.variables.argv.php

Answer 2

Utilize $argv that are predefined variables:

<?php

    var_dump($argv);

Command line:

php args.php -email [email protected]

Exit:

array(3) {
  [0]=>
  string(8) "args.php"
  [1]=>
  string(6) "-email"
  [2]=>
  string(31) "[email protected]"
}

_{To documentation is on the website php.net.}

---

If you want to count the number of arguments use $argc, which is also part of predefined variables:

Command line:

php args.php -email [email protected]

Exit:

int(3)

Remarks: so much $argv and $argc need that register_argc_argv is enabled to function and what defines one argument from the other is space.

References:

• $argv - Array of arguments passed to the script
• PHP args - How to read command line Arguments in PHP
• predefined variables
• register_argc_argv