"php, pass parameters from command line to a php script" Answer’s

0

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:Scriptsscript.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.

Wednesday, March 31, 2021
 
joostvandriel
answered 11 Months ago
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