How to display info from and Mysql table using php

Hi,


I need to pull out some info from a table in my data base using php.

<?php
$result = mysql_query("SELECT 'giftlog.name', 'giftlog.image' FROM giftlog")





LEFT JOIN    gifts

USING (giftid)

WHERE    giftlog.recipientid = " . $userid"

?>

PHP:

This is the code but it doesnt work.....lim a noob at the whole sql/php thing.

 

Thats because you have part of the MySQL query outside of the section that is being queried.

 

You must use following code

$result = mysql_query("SELECT 'giftlog.name' AS name, 'giftlog.image' AS image FROM giftlog LEFT JOIN gifts USING (giftid) WHERE giftlog.recipientid = " . $userid);

$row = mysql_fetch_assoc($result));

echo $row['name'];
echo $row['image'];

PHP:

 

I get this -



Parse error: parse error, unexpected ')' in /home/p/i/picturest/public_html/mygifts.php on line 5

 

Any1 got any idea's

 

Sorry, I put redundant bracket.

Correct line is:

$row = mysql_fetch_assoc($result);

PHP:

 

Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result

 

there is a problem with your $result

try this

$result = mysql_query("SELECT `giftlog`.`name` AS name, `giftlog`.`image` AS image FROM `giftlog` LEFT JOIN `gifts` ON `giftlog`.`giftid` = `gifts`.`giftid` WHERE `giftlog`.`recipientid` = " . $userid);

PHP:

However, i see you dont' pull any fields from the gifts table, so i can't see why you are using the LEFT JOIN

 

ok it displays nothing --which is better then an error !

 

If you can provide access to your database for example via phpMyAdmin maybe I will be able to help you.