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phparrayspopen

in_array() expects parameter 2 to be array, null given in

I'm trying to run a command that outputs when the results whenever something changes. However, I keep getting an in_array() expects parameter 2 to be array, resource given in error.

Here is my code:

$call = "~/www/reteps.tk/go/kahoot-auto " . '262671' . " " . 'bob' . " ";
$result = array();
$old_result = array(0 => "something");

$handle = popen($call, "r");
$result = file($handle); #is this how I read the results?
while (in_array("end",$result) != true) {
  sleep(1); 
  if ($result != $old_result) {
    echo($result);
    $old_result = $result;
  }   
}

Full Error log:

PHP Warning:  file() expects parameter 1 to be a valid path, resource given in /home/retep/www/reteps.tk/school/kahoot2.php on line 70
PHP Warning:  in_array() expects parameter 2 to be array, null given in /home/retep/www/reteps.tk/school/kahoot2.php on line 71
PHP Warning:  in_array() expects parameter 2 to be array, null given in /home/retep/www/reteps.tk/school/kahoot2.php on line 71
PHP Warning:  in_array() expects parameter 2 to be array, null given in /home/retep/www/reteps.tk/school/kahoot2.php on line 71
Solution

  • Can you provide the value of $call? The error seems to indicate that it's a resource instead of a path. popen() requires a $command:

    http://php.net/manual/en/function.popen.php

    So $call should be a string of some kind, either a Linux command or executable file path or something. Right now $call appears to be a resource, so the path or command may be INSIDE of the resource and you just need to point to that value instead of the resource itself.