I am stumped. I am changing my host today and 90% of my sites are all database. I can't seem to query the new database the right way.

This script works fine on the old server but not on the new one.

here are links to the old site and new site...same script

[edit=Republikin]Please no links to adult material or sites containing links to adult material[/edit]

PHP Code:

<?php

include ("include/header.inc.php");

include("include/dbconnect.php");

$result = mysql_query("SELECT * FROM '$table' ORDER BY 'artist'",$db);


$sql = 'SELECT * FROM `musictopten` LIMIT 0, 30 ';
echo "<table cellpadding=3 cellspacing=2 border=0 align=center width=100%><tr><td bgcolor=#000000><b>Artist Name</td><td bgcolor=#000000><b>Feature Video Title</td><td bgcolor=#000000><b>Video</td></tr>"; 

$alternate = "2"; 
while ($row = @mysql_fetch_array($result)) { 
$id = $row["id"]; 
$artist = $row["artist"];
$song1 = $row["song1"];

if ($alternate == "1") { 
$color = "#555555"; 
$alternate = "2"; 
} 
else { 
$color = "#222222"; 
$alternate = "1"; 
} 
echo "<tr bgcolor=$color><td>$artist</td><td>$song1</td><td><a href='view1.php?id=$id'>See All Videos</a></td></tr>"; 
} 
echo "</table>"; 

include ("include/footer.inc.php");
?>

PHP Code:

 <?

include("include/config.php");

$db = mysql_connect("$dbserver", "$dbuser", "$dbpass");
mysql_select_db("$dbname",$db);
  ?>

Not sure what I am doing wrong but I need to get this to work. The new server offers a more updated version of php and mysql so it should work fine and the only part of the code i did change was the @ in front of the query.

Any help would be great and i will give you a credit and a link on my page

I don't honestly see anything wrong with your code. Obviously, if it was working on the old server, then the code works.

Here are some questions to ask yourself. Is the database on the new server using a new name? Is the database even set up? Is the database in place, but each table is empty? Has your username or password changed? Is the "new" version of PHP actually version 5 (version 4 and 5 changed the way to access a database, you might have to rewrite some code)?

i Have changed everything to the new setting but i can't find the problem. I dont know what is up. I am going to look into in more and get back to you..if anyone sees a problem let me know

i also see a poblem with the code posted...i have two querys that are asking the same thing but different ways...the second was a add-on and I took it out. I am still stumprd...I think it is a problem with the way i am calling the tables and not with the way i am calling the database.

run the script with error_reporting(E_ALL); at the top and remove @ sign so that you will be able to know the actual error

I get this error

PHP Code:

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/pappasaa/public_html/nwabars/music_room/top_ten/allentries.php on line 13 


PHP Code:

<?php

include ("include/header.inc.php");

include("include/dbconnect.php");

$result = mysql_query("SELECT * FROM '$table' ORDER BY 'artist'",$db);
error_reporting(E_ALL);

echo "<table cellpadding=3 cellspacing=2 border=0 align=center width=100%><tr><td bgcolor=#000000><b>Artist Name</td><td bgcolor=#000000><b>Feature Video Title</td><td bgcolor=#000000><b>Video</td></tr>"; 

$alternate = "2"; 
while ($row = mysql_fetch_array($result)) { 
$id = $row["id"]; 
$artist = $row["artist"];
$song1 = $row["song1"];

if ($alternate == "1") { 
$color = "#555555"; 
$alternate = "2"; 
} 
else { 
$color = "#222222"; 
$alternate = "1"; 
} 
echo "<tr bgcolor=$color><td>$artist</td><td>$song1</td><td><a href='view1.php?id=$id'>See All Videos</a></td></tr>"; 
} 
echo "</table>"; 

include ("include/footer.inc.php");
?>

You need to move the error reporting up, before the include statements, so you can see the errors with those, too. Sounds like your MySQL isn't set up right.

-T

Try changing this...

PHP Code:

$result = mysql_query("SELECT * FROM '$table' ORDER BY 'artist'",$db); 


To this..

PHP Code:

$result = mysql_query("SELECT * FROM $table ORDER BY 'artist'",$db); 


That's most likely the problem, although if you have any more trouble, add an or die() to the end of the query:

PHP Code:

$result = mysql_query("SELECT * FROM $table ORDER BY 'artist'",$db) or die(mysql_error()); 


That way you can see the mysql errors as well as the php ones. The 'invalid MySQL result resource' error usually means your query failed.