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I am using the Amazon S3 API to upload files and I am changing the name of the file each time I upload. So for example: Dog.png > 3Sf5f.png Now I got the random part working as such: function rand_string( $length ) { $chars = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789"; $size = strlen( $chars ); for( $i = 0; $i < $length; $i++ ) { $str .= $chars[ rand( 0, $size - 1 ) ]; } return $str; } So I set the random_string to the name parameter as such: $params->key = rand_string(5); Now my problem is that this wont show any extension. So the file will upload as 3Sf5f instead of 3Sf5f.png . The variable $filename gives me the full name of the file with its extension. If I use $params->key = rand_string(5).'${filename}'; I get: 3Sf5fDog.png So I tried to retrieve the $filename extension and apply it. I tried more than 30 methods without any positive one. For example I tried the $path_info(), I tried substr(strrchr($file_name,'.'),1); any many more. All of them give me either 3Sf5fDog.png or just 3Sf5f . An example of what I tried: // As @jcinacio pointed out. Change this to: // // $file_name = "${filename}"; // $file_name = '${filename}' // Is is wrong, since '..' does not evaluate $params->key = rand_string(5).$file_name; = 3Sf5fDog.png . $file_name = substr(strrchr('${filename}', '.'), 1); $params->key = rand_string(5).$file_name; = 3Sf5f . $filename = "example.png" // If I declare my own the filename it works. $file_name = substr(strrchr('${filename}', '.'), 1); $params->key = rand_string(5).$file_name; = 3Sf5f.png T

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