jim.davidson Posted May 24, 2010 Share Posted May 24, 2010 I'm trying to populate a dropdown menu from the names of picture files in a directory. I'm following some code from a book and it's not working. I think it because of the variable $theFolder (line 98 of code). Could someone tell me why this code doesn't work? [attachment deleted by admin] Link to comment https://forums.phpfreaks.com/topic/202734-need-some-help-with-build-array-of-a-folder/ Share on other sites More sharing options...
foxsoup Posted May 24, 2010 Share Posted May 24, 2010 It's possibly failing because the directory doesn't exist ('../Misc' as set on line 72), but hard to tell for sure without knowing the exact error message. If you just want a simple bit of code to scan a directory on the server for image files then this should do the trick though: <?php $dirToScan = '.'; // Directory to scan (current dir) $imgExts = array('jpg', 'jpeg', 'png', 'gif', 'bmp'); // Array of possibly image file extensions $imgFiles = array(); // Create blank results array $allFiles = scandir($dirToScan); // Get all files in dir into an array foreach ($allFiles as $filename) { // Loop through each $fileInfo = pathinfo($filename); // Get the info of each file if (in_array($fileInfo['extension'], $imgExts)) { // Check to see if file ext is in array $imgFiles[] = $filename; // If so, add the filename to results array } } var_dump($imgFiles); // Print the results ?> Link to comment https://forums.phpfreaks.com/topic/202734-need-some-help-with-build-array-of-a-folder/#findComment-1062602 Share on other sites More sharing options...
jim.davidson Posted May 24, 2010 Author Share Posted May 24, 2010 My problem was the way I was describing the folder name. i.e. '../Misc/' Once I looked at your example and chaged my folder name to 'Misc' it works. Thank You for your help Link to comment https://forums.phpfreaks.com/topic/202734-need-some-help-with-build-array-of-a-folder/#findComment-1062612 Share on other sites More sharing options...
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