How to retrieve multiple images from MySQL using PHP?

Question

I am able to retrieve single image from mysql database using php.

Now I want to retrieve multiple images.

Here is the piece of code:

<?php
 --db connection code--
 $sql = "SELECT image FROM try WHERE status='0'";
 $result = mysql_query("$sql") or die("Invalid query: " . mysql_error());
 while ($row = @mysql_fetch_assoc($result))
 {
   // set the header for the image
   header("Content-type: image/jpeg");
   echo mysql_result($result, 0);
   echo '<br>';
 }
?>

executing this code, only first image in table is being displayed. How to retrieve multiple images?

You can't show multiple files within one PHP file with header("Content-type: image/jpeg"); — Joran Den Houting
– Joran Den Houting, Commented Sep 20, 2013 at 8:58
first image is displayed because of mysql_result($result, 0); — M Khalid Junaid
– M Khalid Junaid, Commented Sep 20, 2013 at 8:59
When you request an url form a web server you normally request only one file, because your header is just Content-type: image/jpeg the result is also just one file. While multipart responses exist, i currently don't know how well supported they are. — t.niese
– t.niese, Commented Sep 20, 2013 at 8:59

SztupY · Accepted Answer · 2013-09-20 09:13:42Z

1

You are using Content-type: image/jpeg, which means it can only return one image, as the browser will only expect one. If you want to return multiple images to download you have to do some workaround.

One option would be to first pack the files together into one file (like a ZIP archive), and push that to the user.

Or if you don't want to push this as a download, then create a HTML page, with image links to your images. Each of the link will only return one of course.

edited Sep 20, 2013 at 9:13

answered Sep 20, 2013 at 8:59

SztupY

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Comments

M Khalid Junaid · Accepted Answer · 2013-09-20 09:07:40Z

0

You need to pass the loop counter in result_mysql

$count=0;
 while ($row = @mysql_fetch_assoc($result))
 {
   // set the header for the image
   header("Content-type: image/jpeg");
   echo mysql_result($result,$count);

   $count++;
 }

Please don't use the error suppressor @

mysql_result

NOTE : Please don't use mysql* functions in your new code they are depriciated in newer versions use mysqli* or PDO

edited Sep 20, 2013 at 9:07

answered Sep 20, 2013 at 9:01

M Khalid Junaid

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4 Comments

chirag ode Over a year ago

mysql() is deprecated

t.niese Over a year ago

echo <br>; and header("Content-type: image/jpeg"); are you kidding?

M Khalid Junaid Over a year ago

@t.niese i tried to point mysql_result($result, 0); mistake there are also others

t.niese Over a year ago

@dianuj sure, but if you post an answer what should be changed and copying the invalid code as is and just fixing one part, it would be misleading also for others reading this answer. I think it is better to leave the whole invalid code block out an just state what needs to be changed. Or show a correct code block.

Bacis · Accepted Answer · 2013-09-20 09:32:16Z

0

As the Content-type: image/jpeg returns only a single image, you should consider saving the paths of the images in your database table. Later on you won't have any problems fetching all of them and echoing the source elements in your html.

answered Sep 20, 2013 at 9:32

Bacis

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