PHP MySql Picture Data - php

Question

I cannot display an image from my Mysql Database.
I've kept following all the examples and other posts but can't seem to get it right...
<?php
$db = new PDO('mysql:host=localhost;dbname=MyDatabase;charset=utf8mb4', 'tester', '1234567890');
$db->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
$db->setAttribute(PDO::ATTR_EMULATE_PREPARES, false);
$request = "347";
$mytable = "u55";
$stmt = $db->prepare("SELECT * FROM ".$mytable." WHERE Id = :SearchName ");
$stmt->bindParam(':SearchName', $request, PDO::PARAM_STR);
$stmt->execute();
$result = $stmt->fetchAll(PDO::FETCH_ASSOC);
$result = $result[0]['TheAvatar'];
header("Content-Type: image/jpg");
echo '<img src="'.$result.'" />';
?>
My error that I have is image cannot be displayed because it contains errors.
The only difference I have from the examples is the data is stored in TEXT not BLOB/VARCHAR..
I've also tried using BinToHex() with the data to see if that made a difference.
The pictures come from a phone and are stored on a server.
The Pic data is in this link below (cant copy paste it so..)
http://spyon.agency/PicData.bmp
EDIT:
This is the code that stores the data into the DB..
It's a code snippet from the main file, the login remains the same as posted above. The Data comes from a JSON string.
$MessageDecoded = base64_decode ($request);
$input = json_decode($MessageDecoded);
Then the function that stores the data is
$mTheAvatar = $input[$i]->TheAvatar;
$mTheDirection= $input[$i]->TheDirection;
$mTheGroup = $input[$i]->TheGroup;
$mTheMedia = $input[$i]->TheMedia;
$mTheMessage = $input[$i]->TheMessage;
$mTheSenderName= $input[$i]->TheSenderName;
$mTheThumbImage = $input[$i]->TheThumbImage;
$mTheTime = $input[$i]->TheTime;
$mTheMediaExtension = $input[$i]->TheMediaExtension;
$statement = $db->prepare('INSERT INTO '.$mDevice.' '.
'(TheAvatar , TheDirection , TheGroup , TheMedia , TheMediaExtension , TheMessage , TheSenderName , TheThumbImage , TheTime) '.
'VALUES (:aTheAvatar, :aTheDirection, :aTheGroup , :aTheMedia, :aTheMediaExtension , :aTheMessage, :aTheSenderName, :aTheThumbImage, :aTheTime)');
try {
$statement->execute(array(
"aTheAvatar" => $mTheAvatar,
"aTheDirection" => $mTheDirection,
"aTheGroup" => $mTheGroup,
"aTheMedia" => $mTheMedia,
"aTheMediaExtension" => $mTheMediaExtension,
"aTheMessage" => $mTheMessage,
"aTheSenderName" => $mTheSenderName,
"aTheThumbImage" => $mTheThumbImage,
"aTheTime" => $mTheTime
));
} catch(PDOException $ex) {
echo "An Error occured!";
echo $ex->getMessage();
die;
}

Answer 1

you can use this code to show your image :
echo '<img src="data:image/jpeg;base64,'.base64_encode( $result ).'"/>';