how to upload image from database using php mysql code example

Example 1: image upload in php code with databases

<?php
  // Create database connection
  $db = mysqli_connect("localhost", "root", "", "image_upload");

  // Initialize message variable
  $msg = "";

  // If upload button is clicked ...
  if (isset($_POST['upload'])) {
  	// Get image name
  	$image = $_FILES['image']['name'];
  	// Get text
  	$image_text = mysqli_real_escape_string($db, $_POST['image_text']);

  	// image file directory
  	$target = "images/".basename($image);

  	$sql = "INSERT INTO images (image, image_text) VALUES ('$image', '$image_text')";
  	// execute query
  	mysqli_query($db, $sql);

  	if (move_uploaded_file($_FILES['image']['tmp_name'], $target)) {
  		$msg = "Image uploaded successfully";
  	}else{
  		$msg = "Failed to upload image";
  	}
  }
  $result = mysqli_query($db, "SELECT * FROM images");
?>
<!DOCTYPE html>
<html>
<head>
<title>Image Upload</title>
<style type="text/css">
   #content{
   	width: 50%;
   	margin: 20px auto;
   	border: 1px solid #cbcbcb;
   }
   form{
   	width: 50%;
   	margin: 20px auto;
   }
   form div{
   	margin-top: 5px;
   }
   #img_div{
   	width: 80%;
   	padding: 5px;
   	margin: 15px auto;
   	border: 1px solid #cbcbcb;
   }
   #img_div:after{
   	content: "";
   	display: block;
   	clear: both;
   }
   img{
   	float: left;
   	margin: 5px;
   	width: 300px;
   	height: 140px;
   }
</style>
</head>
<body>
<div id="content">
  <?php
    while ($row = mysqli_fetch_array($result)) {
      echo "<div id='img_div'>";
      	echo "<img src='images/".$row['image']."' >";
      	echo "<p>".$row['image_text']."</p>";
      echo "</div>";
    }
  ?>
  <form method="POST" action="index.php" enctype="multipart/form-data">
  	<input type="hidden" name="size" value="1000000">
  	<div>
  	  <input type="file" name="image">
  	</div>
  	<div>
      <textarea 
      	id="text" 
      	cols="40" 
      	rows="4" 
      	name="image_text" 
      	placeholder="Say something about this image..."></textarea>
  	</div>
  	<div>
  		<button type="submit" name="upload">POST</button>
  	</div>
  </form>
</div>
</body>
</html>

Example 2: download image from mysql using php

//You can save this to test.php and call it with test.php?id=1 for example
<?php

//Database class with PDO (MySQL/MariaDB)
require_once("database.php"); //If you need this, write to [email protected] i'll send you the db class

//Connect to database
$database = new Database();
$pdo = $database->dbConnection();

//Get ID from GET (better POST but for easy debug...)
if (isset($_GET["id"])) {
	$id=(int)$_GET["id"];
} else {
  echo "Wrong input";
  exit;
}

//Prepare PDO SQL
$q = $pdo->prepare("SELECT * FROM `table_with_image` WHERE `id`=:p_id");

//Add some data
$q->bindparam(":p_id", $id); //Filter user input, no sanitize necessary here

//Do the db query
$q->execute();

//If something found (always only 1 record!)
if ($q->rowCount() == 1) {
  
  	//Get the content of the record into $row
	$row = $q->fetch(PDO::FETCH_ASSOC); //Everything with id=$id should be in record buffer
  	
  	//This is the image blob mysql item  
  	$image = $row['image'];
  	
  	//Clean disconnect
  	$database->disconnect();
    
  	//Now start the header, caution: do not output any other header or other data!
  	header("Content-type: image/jpeg");
    header('Content-Disposition: attachment; filename="table_with_image_image'.$id.'.jpg"');
    header("Content-Transfer-Encoding: binary"); 
    header('Expires: 0');
    header('Pragma: no-cache');
    header("Content-Length: ".strlen($image));
    //Output plain image from db
	echo $image;
} else {
  //Nothing found with that id, output some error
  $database->disconnect();
  echo "No image found";
}

//No output and exceution further this point
exit();

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