Database connect problem

Tarman · **Database connect problem -** 07-16-2008, 08:05 AM

I keep getting this error when trying to connect to my database any assistance would be helpful.

This is the database connect function

Code:

<?
function database_connect()
{
$dbhost     = "mysql1.000webhost.com";					// Database host
$dbname     = "a1863106_items";			// Database name
$dbusername = "a1863106_tar";			// Database user name
$dbuserpw   = "hidden";						// Database password

mysql_pconnect($dbhost, $dbusername, $dbuserpw);  	//Connects to the mysql database area
$result = mysql_select_db($dbname);   //Connects to the specific database and saves the result as result.
if (!$result)
	return false;  //Connected
else
	return true;  //not connected
}
?>

This is the webpage

Code:

<?
// start the session 
session_start();
ob_start();
include ('includes/allfunctions.php');
header("Cache-control: private"); //IE 6 Fix 
?>
<title>DDO Items</title>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1">
<body>
<table width="100%" border="1" cellspacing="1" cellpadding="1">
  <tr> 
    <? include ('includes/header.php');
	?>
  </tr>
  <tr> 
    <td height="500">
<?
//Connect to the database all fields are good so far				
	database_connect();
	$sql = "SELECT * FROM 'DDO Items' WHERE 1";
	   while($row = mysql_fetch_array($result)) {
    	    echo '<tr>';
        	echo '<td>'.$row['ItemName'].'</td>';
        	echo '</tr>';
}
?>
	</td>
  </tr>
    <td align="center" valign="bottom"><? include ('includes/footer.php')?>
   </td>
   </table>
</html>

This is the error:
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/a1863106/public_html/gloves.php on line 22

Line 22 from above to help find it easier is:

PHP Code:


			
while($row = mysql_fetch_array($result)) {

Red-X · 07-16-2008, 08:25 AM

Try this <?php not <?. I think that's the error hope that helps

Tarman · 07-16-2008, 02:22 PM

Sorry still no luck any other possible suggestions?

admin · 07-16-2008, 03:30 PM

can you enter database by phpmyadmin? If yes - its all fine with database and problem somewhere with script coding.

Tarman · 07-17-2008, 01:03 AM

Yes I can enter PHP my admin....I understand the database is good I am not sure what I am missing in my PHP.

Once again here is my codeing

Code:

<?
//Connect to the database all fields are good so far				
	database_connect();
	$sql = "SELECT * FROM 'DDO Items' WHERE 1";
	$result = mysql_query($sql);
	   while($row = mysql_fetch_array($result)) {
    	    echo '<tr>';
        	echo '<td>'.$row['ItemName'].'</td>';
        	echo '</tr>';
}
?>

Red-X · 07-17-2008, 01:11 AM

Try this..

PHP Code:


			
echo "<td>" . $row["ItemName"] . "</td>";

Mr.Whisper · 07-18-2008, 07:39 AM

I usually use this

Code:

<?
$sql = "SELECT * FROM my_table WHERE condition=condition";
$query = mysql_query($sql);
$rows = mysql_num_rows($query);
$i = 0;
while($i < $rows){
	$result = mysql_fetch_array($query);
	echo $rows['field'];
	$i++;
}
?>

so, at first you was lost query function(you forgot to defing $result) and then on your new code. it still show warning? so i think my code will work fine.

i always use basic syntax. i tried to fix problem that occur with advance syntax

Tarman · 07-18-2008, 02:30 PM

Thanks for all the help and suggestions. I got it to work Here is the code if it could help somebody else.

Code:

<?
//Connect to the database all fields are good so far

$sql = 'SELECT * FROM `a1863106_items`.`DDO Items` WHERE `ItemType` LIKE CONVERT(_utf8 \'%Armor%\' USING latin1) COLLATE latin1_general_ci'; 		
	database_connect();
	$result = mysql_query($sql) or die(mysql_error());
	   while($row = mysql_fetch_array($result)) {
			echo '<tr>';
        	echo '<td>'.$row['ItemName'].'</td>';
			echo '<td>'.$row['ItemDescription'].'</td>';
			echo '<td>'.$row['Type'].'</td>';
			echo '<td>'.$row['QuestName'].'</td>';
			echo '<td>'.$row['lvl'].'</td>';		
			echo '<td>'.$row['Notes'].'</td>';
        	echo '</tr>';
}

$sql = 'SELECT * FROM `a1863106_items`.`DDO Items` WHERE `ItemType` LIKE CONVERT(_utf8 \'%Docent%\' USING latin1) COLLATE latin1_general_ci'; 		
	database_connect();
	$result = mysql_query($sql) or die(mysql_error());
	   while($row = mysql_fetch_array($result)) {
			echo '<tr>';
        	echo '<td>'.$row['ItemName'].'</td>';
			echo '<td>'.$row['ItemDescription'].'</td>';
			echo '<td>'.$row['Type'].'</td>';
			echo '<td>'.$row['QuestName'].'</td>';
			echo '<td>'.$row['lvl'].'</td>';		
			echo '<td>'.$row['Notes'].'</td>';
        	echo '</tr>';
}


?>

KazanDaemon · 08-20-2008, 05:21 PM

Quote:

Originally Posted by Tarman

Yes I can enter PHP my admin....I understand the database is good I am not sure what I am missing in my PHP.

Once again here is my codeing

Code:

<?
//Connect to the database all fields are good so far				
	database_connect();
	$sql = "SELECT * FROM 'DDO Items' WHERE 1";
	$result = mysql_query($sql);
	   while($row = mysql_fetch_array($result)) {
    	    echo '<tr>';
        	echo '<td>'.$row['ItemName'].'</td>';
        	echo '</tr>';
}
?>

try ...FROM `DDO Items` WHERE...
database field names MUST BE quoted by (`) char, not (')

YaThisHostRox · 08-24-2008, 10:43 PM

Use ` not '. I make that mistake muchos.