php - Storing images as a blob in mysql database -


so have searched high , low relevant answer, , have yet find one, i'm going ask question myself. in mysql tables have 5 columns of id, customer__id, file_data, file_name, mime_typethe relevant 1 here file_data type of longblob. understand size of files longblob can handle pretty substantial when try upload file around 978 kb fails, dimensions of image large (2048 x 1536)?

here code uploader. works elegantly things excel sheets, pdf, word documents, , other stuff, when comes images fails:

<?php      require_once '../../inc/config.php';     $response = array();     $response['errors'] = false;     $id = $_request['id'];      if(!empty($_files)){         //set default data arrays         $names = array(); //stores file names         $files = array(); //stores file data         $mime_types = array(); //store file type mime type          //force each file name names array         foreach($_files['file']['name'] $name){             array_push($names, $name);         }         //force file data own array spot in files array         foreach($_files['file']['tmp_name'] $temp){             array_push($files, prepareimagedbstring($temp));         }         //force mimetypes mime_types array         foreach($_files['file']['type'] $type){             array_push($mime_types, $type);         }          //process 3 of file arrays simultaneously no data left out         for($i = 0; $i < count($names); $i++){             $file_name = $names[$i];             $file_data = $files[$i];             $mime_type = $mime_types[$i];              //set query data go note_file table in database              $q = "insert brb.files (customer__id, file_name, file_data, mime_type)             values('$id', '$file_name', '$file_data', '$mime_type')";             //run query             if($stmt = $conn->prepare($q)){             //process errors may occur                  if(!$stmt->execute()){                     printf("error message: %s\n", $conn->error);                 }             }         }      }       echo json_encode($response);      function prepareimagedbstring($filepath){         $out = 'null';         $handle = fopen($filepath, 'r');         if($handle){             $content = fread($handle, filesize($filepath));             $content = bin2hex($content);             fclose($handle);             $out = $content;         }          return $out;     } ?> 

if point me in direction fantastic. please not provide answer on why practice bad, i'm aware it's bad practice , know proper practice is, learning tool myself , nothing more, bear me on minute.

thank provides help!