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I created a MySQL database with a table using phpmyadmin. I created this table with a BLOB column to hold a jpeg file.

I have issues with regards to the php variable $result here.

My code so far: (catalog.php):

<body> <?php $link = mysql_connect("localhost", "root", ""); mysql_select_db("dvddb"); $sql = "SELECT dvdimage FROM dvd WHERE id=1"; $result = mysql_query("$sql"); mysql_close($link); ?> <img src="" width="175" height="200" /> </body>

How can I get the variable $result from PHP into the HTML so I can display it in the <img> tag?

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6
  • 1
    Whaaaat issues? You need to say what your code is doing wrong at the moment. Commented Oct 17, 2011 at 11:22
  • 1
    You can start by fetching those results Commented Oct 17, 2011 at 11:23
  • i am not clear of the syntax involved to display the BLOB image Commented Oct 17, 2011 at 11:25
  • @DamienPirsy How can i fetch these results? Commented Oct 17, 2011 at 11:26
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    Just my 2 cents and not related to your question, but I think you'd be better off storing the images on your file system and just referencing the path in your database. You'll get better performance, and more flexibility as to where you store your images. Commented Oct 17, 2011 at 11:29

5 Answers 5

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You can't. You need to create another php script to return the image data, e.g. getImage.php. Change catalog.php to:

<body> <img src="getImage.php?id=1" width="175" height="200" /> </body>

Then getImage.php is

<?php $link = mysqli_connect("localhost", "root", "", "dvddb"); $sql = "SELECT dvdimage FROM dvd WHERE id=?"; $result = mysqli_execute_query($link, $sql, [$_GET['id']]); $image = mysqli_fetch_column($result); header("Content-type: image/jpeg"); echo $image;
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7 Comments

but the thing is,I have nearly 50 images in my db.You have written for getting only one image.How can i display whole images in the page
To get a different image from the DB you simply change the id get parameter: <img src="getImage.php?id=1" width="175" height="200" /> <img src="getImage.php?id=2" width="175" height="200" /> ...
This answer is incorrect in that it IS possible, view Ilmari's answer.
@DheerajMPai That is the entire script. Although these days you'd use MySQLi instead of the mysql_* functions (this answer is 7 years old).
It may be 7 year old but still helpful for beginners. Anyway I shifted for python. So no issues anymore!@megaflop
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Technically, you can too put image data in an img tag, using data URIs.

<img src="data:image/jpeg;base64,<?php echo base64_encode( $image_data ); ?>" />

There are some special circumstances where this could even be useful, although in most cases you're better off serving the image through a separate script like daiscog suggests.

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1 Comment

I was wondering whether someone would suggest a data URI. Not very efficient, IMO (no browser/proxy caching), but certainly worthy of a note.
4

In case you store only the filename, while the image itself is stored in a file

while($row = mysqli_fetch_array($result)) { echo 'img src="',htmlspecialchars($row['filename']),'" width="175" height="200" />"; }
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0

Using PDO

Query

<?php require 'connect.php'; $stmt = $pdo->query("SELECT * FROM products"); $products = $stmt->fetchAll(PDO::FETCH_ASSOC);

View

<table> <thead> <tr> ... <th>Product Name</th> <th>Product Image</th> </tr> </thead> <tbody> <?php foreach ($products as $product) : ?> <tr> ... <td><?= htmlspecialchars($product['product_name']); ?></td> <td> <img src="data:image/jpeg;base64,<?= base64_encode($product['image']); ?>" width="100" height="100" > </td> </tr> <?php endforeach; ?> </tbody> </table>
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-2

add $row = mysql_fetch_object($result); after your mysql_query();

your html <img src="<?php echo $row->dvdimage; ?>" width="175" height="200" />

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1 Comment

You can't put raw image data inside an img src. OP clearly states the database stores the image data, not a path to the image file.

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