Trying to echo an Image from BLOB

Question

I have a page where a user enters an ID for a book record and it is supposed to pull the corresponding image (only the image) that was entered with a books details from the database which is stored as a BLOB. - Yes, its better to store the image file path - I know this, but that's not what i'm trying to achieve here.

Currently when I press the submit button to find an ID specific by user it displays a page with an image symbol, but no image.

Here is my code.

require_once __DIR__.'/config.php';
session_start();
$dbh = new PDO('mysql:host=' . DB_HOST . ';dbname=' . DB_USERNAME, DB_USERNAME, DB_PASSWORD);

$book = $_POST["book_id"];
$stmt = $dbh->prepare("select image from books2 where b_id = $book ");
$stmt->execute();


if ($stmt) {
    if ($row = $dbh->prepare($stmt)) {
       $img = $row["image"];
    }
}

header("Content-type: image/jpeg");

echo "$img";

How do I get the image to display, I have read other questions but haven't been able to figure it out?

Answer 1

You have an sql injection problem and you are not fetching a row with the image data:

$book = $_POST["book_id"];
$stmt = $dbh->prepare("select image from books2 where b_id = $book ");
$stmt->execute();


if ($stmt) {
    if ($row = $dbh->prepare($stmt)) {
       $img = $row["image"];
    }
}

Should be something like:

$stmt = $dbh->prepare("select image from books2 where b_id = ?");
$stmt->execute(array($_POST["book_id"]));

if ($row = $stmt->fetch()) {
   $img = $row["image"];
}

Also note that you don't have any error handling in case no row is found. You should add that as well.

Answer 2

Instead of adding like this its better to do using the below method.

Just add your code in one file lets call display.php

require_once __DIR__.'/config.php';
session_start();
$dbh = new PDO('mysql:host=' . DB_HOST . ';dbname=' . DB_USERNAME, DB_USERNAME, DB_PASSWORD);

$book = $_GET["book_id"];
$stmt = $dbh->prepare("select image from books2 where b_id = $book ");
$stmt->execute();


if ($stmt) {
    if ($row = $dbh->prepare($stmt)) {
       $img = $row["image"];
    }
}

header("Content-type: image/jpeg");

echo "$img";

Then using the img tag you can easily print the image by passing the book_id as a url in src

As like below

<img src="display.php?book_id=123" alt="your alt title"  />

Answer 3

This is the working code thanks to the help of this question : PHP: Retrieve image from MySQL using PDO suggested by Ryan Vincent.

$book = $_GET["book_id"];
$sql = "SELECT image FROM books2 WHERE b_id=:id";
$query = $dbh->prepare($sql);
$query->execute(array(':id' => $book));

$query->bindColumn(1, $image, PDO::PARAM_LOB);
$query->fetch(PDO::FETCH_BOUND);
header("Content-Type: image");
echo $image;