php image blob wont show

Question

So I am trying to show an image from my mysql database. The image is saved as a blob format. I made seperate php file for fetching the image. I then call this file from where I want to show the image like so:

<img src="image.php?id=3" />

This is the file that fetches the image:

<?php
    require 'scripts/connect.php';

    if(isset($_GET['id']))
    {
        $id = mysql_real_escape_string($_GET['id']);
        $q = $db->prepare("SELECT image FROM products WHERE id = '$id'");
        $q->execute();
        $data = $q->fetch();

        header("content-type: image/jpeg");
        echo $data["image"];
    }
?>

But when I try and run this code no image shows up. Instead I get this annoying broken link image thing:

https://i.sstatic.net/LGGTF.png

What is size of image? did you try to open image file from some (text/hex) editor? — Nebojsa Susic
– Nebojsa Susic, Commented Jul 8, 2014 at 12:45
I think you have a misunderstanding on how to use prepared statements. Use placeholders in your SQL that will be prepared, don't use user input there. User input should go to execute(). — vhu
– vhu, Commented Jul 8, 2014 at 12:49
I am sure the image is of jpeg format and yes I just face palmed over my prepared statement. — user3511194
– user3511194, Commented Jul 8, 2014 at 12:59

DonCallisto · Accepted Answer · 2014-07-08 13:35:09Z

1

Your code doesn't do what you expect.

Try to change

$q = $db->prepare("SELECT image FROM products WHERE id = '$id'");

in - if id field is numeric one; if isn't, add single quote -

$q = $db->prepare("SELECT image FROM products WHERE id = $id");

Your example didn't work as you were passing to query $id placeholder and not his value (you dind't concatenated it)

Of course with that method you're not save by SQL Injection at all, so you should use pepared statement that way

$q = $db->prepare("SELECT image FROM products WHERE id = :id");
$q->execute(Array(":id" => $id));

Edit

As OP told me that $data['image']; is a bitstream, I will suggest to use something like:

echo '<img src="data:image/jpg;base64,'. base64_encode($data['image']). '" alt='imagename'>;

or if your echo goes directly into src attribute:

echo 'data:image/jpg;base64,'. base64_encode($data['image'])

edited Jul 8, 2014 at 13:35

answered Jul 8, 2014 at 12:43

DonCallisto

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9 Comments

vhu Over a year ago

I'd suggest that you use placeholders instead of entering user input into the SQL statement that will be prepared. No matter if it is escaped or not. Therefore change it to $q = $db->prepare("SELECT image FROM products WHERE id = :id"); and $q->execute(Array(":id" => $id));

DonCallisto Over a year ago

@vhu: you're totally right, I was only suggesting to OP how to make his code work ... I was integrating my answer while my phone start ringing :)

user3511194 Over a year ago

Thanks for you supreme answer. I tried the fix, but the same error occurs.

DonCallisto Over a year ago

@user3511194: we should know how the url is encoded into db ... maybe is a relative one so your "echo" couldn't be sufficient there ...

user3511194 Over a year ago

Okay. How can I see the url?

|

Samuil Banti · Accepted Answer · 2014-07-08 12:45:04Z

0

Try to replace

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

with

header("content-type: image/png");

answered Jul 8, 2014 at 12:45

Samuil Banti

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Comments

Swaraj Giri · Accepted Answer · 2014-07-08 12:58:07Z

0

Try,

$query = $db->prepare("SELECT image FROM products WHERE id = :id");
$query->execute(array(":id" => $id));
$data = $q->fetch();

For serving the image, use

$mime = pathinfo($data['image'], PATHINFO_EXTENSION);
header('Content-Type: '.$mime);
ob_clean();
flush();
readfile($data['image']);

Note:

readfile() needs the image path to where the images are stored.
if you are use PHP >= 5.4, $query->execute([":id" => $id]); can be used instead of $query->execute(array(":id" => $id));

answered Jul 8, 2014 at 12:58

Swaraj Giri

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2 Comments

user3511194 Over a year ago

This didnt work for me. I got the error: Warning: readfile() expects parameter 1 to be a valid path, string given in C:\xampp\htdocs\haugaard\image.php on line 17

Swaraj Giri Over a year ago

that's because readfile needs the exact path to the location of the image.

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