[SOLVED] mysql help ..

[imarockstar]

I am gettng an error with this query ...

 

$sql="SELECT tourdates.id AS tdid, bands.id AS bandid, * FROM tourdates INNER JOIN bands ON tourdates.bandid = bands.id WHERE tourdates.id = '$id' && tourdates.userid = ".

 

error :

 

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/tourzag/public_html/display_tourdate.php on line 41

 

 

more code :

 

<?php

$user = $_SESSION['id'];
$id = $_GET['id'];
$bandid = $_GET['bandid'];

// Retrieve data from database 
$sql="SELECT tourdates.id AS tdid, bands.id AS bandid, * FROM tourdates INNER JOIN bands ON tourdates.bandid = bands.id WHERE tourdates.id = '$id' && tourdates.userid = ". $_SESSION['id'];
$result=mysql_query($sql);


// Start looping rows in mysql database.
$rows = mysql_fetch_array($result);   // this is line #41

?>

 

 

 

 

 

 

[Ken2k7]

$result=mysql_query($sql) or die(mysql_error());

Can you put that up and see what error you get back? But I wonder if your alias of bandid can be causing the problem. Either case, this isn't in the right forum.

[imarockstar]

You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '* FROM tourdates INNER JOIN bands ON tourdates.bandid = bands.id WHERE tourdates' at line 1

[Ken2k7]

This is solved?