mysql_fetch_array(): supplied argument is not a valid MySQL

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

is the error I get 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

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 2

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 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.

Answer 4

Probably happen some SQL error, and the variable $result is not one valid resource... use function mysql_error to debug your application...

For render your output of function you can do only this...

$return = array();

while($row = mysql_fetch_array($result)) {
   $return[] = $row;
}

return $return;

I recommend to you try use some framework as Zend_DB or Doctrine :D

Good luck!