mysql_fetch_array(): supplied argument is not a valid MySQL - php

Question

Warning:mysql_fetch_array(): supplied argument is not a valid MySQL result resource in **/home/davzyco1/public_html/notes/functions.php** on line 43
was the error I got when I use the below class, even though the class works PERFECTLY with my old webhost. Here's my new hosts php info: http://davzy.com/notes/php.php
class mysqlDb
{
public $con;
public $debug;
function __construct($host,$username,$password,$database)
{
$this->con = mysql_connect($host,$username,$password);
if (!$this->con)
{
die('Could not connect: ' . mysql_error());
}
mysql_select_db($database, $this->con);
}
function kill()
{
mysql_close($this->con);
}
function debugOn()
{
$this->debug = true;
}
function debugOff()
{
$this->debug = false;
}
function select($query,&$array)
{
$c = 0;
$result = mysql_query("SELECT ".$query);
if($this->debug == true)
echo "SELECT ".$query;
while($row = mysql_fetch_array($result))
{
foreach($row as $id => $value)
{
$array[$c][$id] = $value;
}
$c++;
}
}
function update($update, $where,$array)
{
foreach($array as $id => $value)
{
mysql_query("UPDATE {$update} SET {$id} = '{$value}'
WHERE {$where}");
if($this->debug == true)
echo "UPDATE {$update} SET {$id} = '{$value}'
WHERE {$where}<br><br>";
}
}
function updateModern($update, $where,$array)
{
foreach($array as $id => $value)
{
mysql_query("UPDATE {$update} SET `{$id}` = '{$value}'
WHERE {$where}");
if($this->debug == true)
echo "UPDATE {$update} SET {$id} = '{$value}'
WHERE {$where}<br>";
}
}
function delete($t, $w)
{
mysql_query("DELETE FROM `{$t}` WHERE {$w}");
if($this->debug == true)
echo "DELETE FROM `{$t}` WHERE {$w}<br><br>";
}
function insert($where, $array)
{
$sql = "INSERT INTO `{$where}` (";
$sql2 = " VALUES (";
foreach($array as $id => $value){
$sql .= "`{$id}`, ";
$sql2 .= "'{$value}', ";
}
mysql_query(str_replace(', )',')',$sql.")") . str_replace(', )',')',$sql2.");"));
if($this->debug == true)
echo str_replace(', )',')',$sql.")") . str_replace(', )',')',$sql2.");")."<br><br>";
}
}

Answer 1

This is because mysql_query() will return FALSE if an error occured, instead of returning a result resource. You can check the error by calling mysql_error(), as shown here:
function select($query,&$array)
{
$c = 0;
$result = mysql_query("SELECT ".$query);
if($this->debug == true)
echo "SELECT ".$query;
if (!$result) {
// an error occured, let's see what it was
die(mysql_error());
}
while($row = mysql_fetch_array($result))
{
foreach($row as $id => $value)
{
$array[$c][$id] = $value;
}
$c++;
}
}
Based on the error message, you can find out what the real problem is.

Answer 2

You should really test to see if $result is not false before using it with mysql_fetch_array. The error you're receiving is indicative that the query itself failed.
Have you configured your database with your new host? (do all the tables exist?)

Answer 3

As Mark said above, you really should check your result before trying mysql_fetch_array on a result set, and verify that all tables actually exist.
Without knowing how your original server was set up, I can only guess, but it may also be that your old server was set up to not display warnings.