How to pass variable to PHP file in Shell Script

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how can I pass a variable to a php file in a shell script, I’m trying to do this way below, without success.

php -f complete.php?login=$1
php linux shell-script

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There will be a variable $argv that will be a array with values passed via CLI, but not the way you did. Just pass the value as argument from script:

$ php -f complete.php teste

Thus, if you value $argv will be ['teste'].

See more in the official documentation: Array of arguments passed to the script.

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