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Micheal Russell
Courses Plus Student 1,312 Pointsarray name $files
array name $files
<?php
//add code here
$files_array = file($dir);
2 Answers

Umesh Ravji
42,362 PointsHi Micheal, tests tend to be pretty specific, so you have to give them exactly what they are looking for.
- The variable must be named
$files
not$files_array
- The function you are after is
scandir()
, which returns an array of all the files within a directory. Thefile()
function will read the contents of a file into an array. - The directory is not provided as a
$dir
variable, but is a directory namedexample

Micheal Russell
Courses Plus Student 1,312 PointsThank you very much. I did not get that from the video but I understand your point. Thank you