Storing images as a blob in mysql database - php

Question

So I have searched high and low for a relevant answer, and I have yet to find one, so I'm just going to ask the question myself. In my MySQL tables I have 5 columns of id, customer__id, file_data, file_name, mime_typethe relevant one here is file_data which is a type of LONGBLOB. Now from what I understand the size of files that a LONGBLOB can handle is pretty substantial but when I try to upload a file that is around 978 KB it fails, could it be that the dimensions of the image are too large (2048 x 1536)?
Here is the code for my uploader. It works very elegantly for things like excel sheets, pdf, word documents, and other stuff, but when it comes to images it 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 the file data
$mime_types = array(); //store the file type as a mime type
//force each file name to the names array
foreach($_FILES['file']['name'] as $name){
array_push($names, $name);
}
//force the file data into its own array spot in the files array
foreach($_FILES['file']['tmp_name'] as $temp){
array_push($files, prepareImageDBString($temp));
}
//force the mimetypes into the mime_types array
foreach($_FILES['file']['type'] as $type){
array_push($mime_types, $type);
}
//process all three of the file arrays simultaneously so that no data is left out
for($i = 0; $i < count($names); $i++){
$file_name = $names[$i];
$file_data = $files[$i];
$mime_type = $mime_types[$i];
//set the query for the data to go into the note_file table in the database
$q = "INSERT INTO brb.files (customer__id, file_name, file_data, mime_type)
VALUES('$id', '$file_name', '$file_data', '$mime_type')";
//run the query
if($stmt = $CONN->prepare($q)){
//process any errors that 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 someone could point me in the direction that would be fantastic. Please do not provide be an answer on why the practice is bad, I'm aware it's bad practice and know what proper practice is, this is a learning tool for myself and nothing more, just bear with me on it for a minute.
Thank you to everyone who provides help!