Displaying an image in php using mysql

Question

I'm trying to display an image along with other information from the database.

PHP

  <?php
               mysql_connect("localhost","111","222") or die("Could not connect to localhost");
               mysql_select_db("movies") or die( "Could not connect to database");

               $result = mysql_query("SELECT * FROM allmovies");
               if ($result == false )  die(mysql_error()); 
               while($row = mysql_fetch_array($result))
               {
                    echo "<img src=' . $row['image'] . '>";
               ?>

Like:

Title: Blah
Price: Blah
Image: <img src=rarara">

All from MySQL in one page?

I hope you've got a good reason for using the ancient mysql_query interface. — tadman
– tadman, Commented Oct 4, 2013 at 17:40

Sammitch · Accepted Answer · 2013-10-04 17:56:11Z

2

Don't store image data in a database, they are generally not suited to this and incurs extra overhead on your MySQL connections returning the data. You should be storing the path to the image file, and serving that.
If you insist on doing it you you should only be returning one image at a time with the proper header based on the image type using something like the following:
```
$imagedata = data_from_mysql();
header('Content-Length: ' . sizeof($imagedata) );
header('Content-Type: image/png');
echo $imagedata;
exit;
```

If you really want to make your page source bloated, slow, unmanageable, and nigh-uncacheable:

while( $imagedata = data_from_mysql() ) {
  echo "<img src='data:image/png;base64," . base64_encode($imagedata) . "'>";
}

I cannot stress enough how these are terrible ideas that you should not use, but if you cannot listen to reason you can at least do bad things the right way.

answered Oct 4, 2013 at 17:56

Sammitch

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5 Comments

user2804639 Over a year ago

@ Sammitch okay, thanks. I changed it to the path now. I'm trying to use echo "<img src=' . $row['image'] . '>";

Sammitch Over a year ago

You're mixing single and double quotes there, echo '<img src="' . $row['image'] . '" />'; or better yet printf('<img src="%s" />', $row['image']);

Kai Qing Over a year ago

Are you saying you just changed the structure of your site and are now storing paths instead of image blobs? If so, consider relative or absolute paths. <img src="/uploads/images/<?php echo $row['image']; ?>"> where $row['image'] = something like 'entry-1.jpg' as opposed to storing a full path in your database. It makes more sense to store filename, then make the path a system variable in case you move things around

Sammitch Over a year ago

@KaiQing I'd recommend against absolute paths since that ties your code to the particular machine it's on at the time and makes changing hosts/machines/providers/OSes/etc a major pain.

Kai Qing Over a year ago

@Sammitch - I mean writing absolute in the src attr of the tag instead of echoing $row['image'] as is. Assuming the db result is just a file name or something

Kai Qing · Accepted Answer · 2013-10-04 18:02:26Z

1

You could use imagecreatefromstring()

$im = imagecreatefromstring($row['image']);
if ($im !== false) {
    ob_start();
    imagejpeg($im);
    $data = ob_get_contents();
    ob_end_clean();
    echo '<img src="data:image/jpg;base64,' .  base64_encode($data)  . '" />';
}

Just my opinion, but it might be slightly more sane to save the images to the file server and then store a reference to the path instead of the whole image as a blob?

answered Oct 4, 2013 at 18:02

Kai Qing

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