PHP, pass parameters from command line to a PHP script

Question

I want to pass parameters from PHP Command Line Interface, and then read in the values using PHP script, something like this:

<?php
  $name1 = $argv[1];    
  echo $name1;
?>

I pass the variable from CLI like this:

C:\xampp\php\php.exe name.php Robby

The above works, I get Robby as the output.

But I want to do something like this:

C:\xampp\php\php.exe name.php -inputFirstName="Robby"

So that the user is well informed to enter the correct parameters in the correct places. What is the appropriate way to parse these parameters?

Answer 1

When calling a PHP script from the command line you can use $argc to find out how many parameters are passed and $argv to access them. For example running the following script:

<?php
    var_dump($argc); //number of arguments passed 
    var_dump($argv); //the arguments passed
?>

Like this:-

php script.php arg1 arg2 arg3

Will give the following output

int(4)
array(4) {
  [0]=>
  string(21) "d:\Scripts\script.php"
  [1]=>
  string(4) "arg1"
  [2]=>
  string(4) "arg2"
  [3]=>
  string(4) "arg3"
}

See $argv and $argc for further details.

To do what you want, lets say

php script.php arg1=4

You would need to explode the argument on the equals sign:-

list($key, $val) = explode('=', $argv[1]);
var_dump(array($key=>$val));

That way you can have whatever you want in front of the equals sign without having to parse it, just check the key=>value pairs are correct. However, that is all a bit of a waste, just instruct the user on the correct order to pass the arguments.