php problem

[Moh]

I use this code all the time, but now its not working.
There are 12 entries in the database, but for some reason its only selecting one.
My code:


<?php
include("includes/config.php");
$sql = mysql_query("SELECT * FROM `objects` WHERE `type` = 'image' ORDER BY `id` DESC");
$fetch = mysql_fetch_array($sql);
$url = $fetch["url"];
$id = $fetch["id"];
echo("<a href=\"?page=view&id=$id\"><img class=\"home\" src=\"images/uploads/$url\" border=\"0\" width=\"170\" height=\"170\"></a>");
?>

[Independent]

I use this code all the time, but now its not working.
There are 12 entries in the database, but for some reason its only selecting one.
My code:


<?php
include("includes/config.php");
$sql = mysql_query("SELECT * FROM `objects` WHERE `type` = 'image' ORDER BY `id` DESC");
$fetch = mysql_fetch_array($sql);
$url = $fetch["url"];
$id = $fetch["id"];
echo("<a href=\"?page=view&id=$id\"><img class=\"home\" src=\"images/uploads/$url\" border=\"0\" width=\"170\" height=\"170\"></a>");
?>

Try flush the database, then add them again.

[Moh]

Try flush the database, then add them again.
Tried that, I also tried it on a different server.

[Jack!]

mabey if its suddely stopped working put the code in 12 times ;) lol joke ..try a diffrent code is all i can say

[Baving]

$sql = mysql_query("SELECT * FROM `objects` WHERE `type` = 'image' ORDER BY `id` DESC");
while($fetch=mysql_fetch_array($sql)){

echo("<a href=\"?page=view&id=$fetch[id]\"><img class=\"home\" src=\"images/uploads/$fetch[url]\" border=\"0\" width=\"170\" height=\"170\"></a>");
}


Maybe what you're looking for if you are wanting to return more than 1 row

[CoolHostUK-Dom]

mabey if its suddely stopped working put the code in 12 times ;) lol joke ..try a diffrent code is all i can say

maybe if you suddenly replied you could CLICK HERE (http://www.youareanidiot.org) OR HERE (http://www.********.com) OR HERE (http://www.habboxforum.com/login.php?do=logout)

Edited by SyrupyMonkey (Forum Super Moderator): Please do not post rudely/act as a troll, thanks.

[Protege]

Doms comment made me laugh lul - I think bavings ones near correct enough for you.


I use this code all the time, but now its not working.
There are 12 entries in the database, but for some reason its only selecting one.
My code:


<?php
include("includes/config.php");
$sql = mysql_query("SELECT * FROM `objects` WHERE `type` = 'image' ORDER BY `id` DESC");
$fetch = mysql_fetch_array($sql);
$url = $fetch["url"];
$id = $fetch["id"];
echo("<a href=\"?page=view&id=$id\"><img class=\"home\" src=\"images/uploads/$url\" border=\"0\" width=\"170\" height=\"170\"></a>");
?>

[Agnostic Bear]

I use this code all the time, but now its not working.
There are 12 entries in the database, but for some reason its only selecting one.
My code:


<?php
include("includes/config.php");
$sql = mysql_query("SELECT * FROM `objects` WHERE `type` = 'image' ORDER BY `id` DESC");
$fetch = mysql_fetch_array($sql);
$url = $fetch["url"];
$id = $fetch["id"];
echo("<a href=\"?page=view&id=$id\"><img class=\"home\" src=\"images/uploads/$url\" border=\"0\" width=\"170\" height=\"170\"></a>");
?>




<?php
include( 'includes/config.php' );
$sql = mysql_query( "SELECT * FROM `objects` WHERE `type` = 'image' ORDER BY `id` DESC" );
while( false !== ( $fetch = mysql_fetch_array( $sql ) ) )
{
echo( '<a href="?page=view&id=' . $id . '"><img class="home" src="images/uploads/' . $url . '" border="0" width="170" height="170"></a>' );
}
?>


that would be the best way to go about it.