Displaying all images in Database table using PHP

elbow316 · 16-03-2006 04:53PM #1

Hi guys,

This is giving me a lot of problems. I am able to insert in images in a table, and display ONE image. But I want to display all the images in the table, this is me giving trouble, this code at the moment is displaying the same image ten times.

As I am incrementing the $ID variable, I want different images to be displayed, but I just keep getting the same one. I do know it is working to some extent, when I put in this statement(echo $name; ) I do get all the image names.

CODE

<?

$db = mysql_connect("******", "*****","*****");

mysql_select_db("****",$db);

$id = 1;
while($id<10){
$query = "SELECT name, type, size, content FROM upload where id ='$id'";
$result = mysql_query($query) or die('Error, query failed');
list($name, $type, $size, $content) = mysql_fetch_array($result);

if ($_REQUEST[gim] == $id) {

header("Content-length: $size");
header("Content-type: $type");

echo $content;

exit;
}

?>

<html>
<title>Upload an image to a database</title><body>

<center><img src=?gim=<? echo $id ?> width=160 height=160><br>

</body>
</html>

<?
$id++;
}
?>

_______________________________
Edit - I had pasted in the wrong version of the code, which was calling 1 as the value, rather than the incremented $id............this version seems to be calling the correct image address(gim=2 or gim=3 and so on) but isn't displaying the relevant image

rsynnott · 16-03-2006 05:07PM

Well, for a start, don't store images in a table unless you've got a very good reason for it.

frost · 16-03-2006 05:27PM

You aren't incrementing $id in your loop.

timeout · 16-03-2006 06:11PM

You aren't incrementing $id in your loop.

Well its in it now at the end.

This is your problem

where id ='$id'";

should be

where id=" . $id;

OR

where id='" . $id . "'";

if you need the single quotes.

frost · 16-03-2006 06:27PM

timeout wrote:

Well its in it now at the end.

Oops, missed that at the end!

timeout wrote:
This is your problem

should be
where id=" . $id;
OR
where id='" . $id . "'";
if you need the single quotes.

Should also work with curly braces:

$query = "SELECT name, type, size, content FROM upload where id ='{$id}'";

Displaying all images in Database table using PHP

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