I’m trying to execute this code:
function main(){ if ($argc < 1){ listDir("."); } else{ for($i = 0; $i <= sizeof($argv); $i++){ listDir($argv[$i]); } } }
But I’m getting the following error:
PHP Notice: Undefined variable: argc in /home/me/test.php on line 15
I thought that $argv and $argc were global variables. How can I get rid of this error?
I’m running this from command line.
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Answer
Add a
global $argc, $argv;
after
function main() {
Those variables are in global scope, but not in your function’s scope. The global keyword imports them.