Problem:mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource

**tony007** · Apr 27th, 2006, 08:27 PM

Hi all i got a php script that works perfectly in my localhost but when i use it in a site which is in remote i get the following error:

Code:

Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in /home2/method/public_html/music/singerIndex.php on line 72

When i load the page no output and i get the above error. Is there any connection problem here? or sql syntax ?I be happy if an expert tell me how to fix it.Thanks

Code:

<?php


//alone script that reads folder but but not subfolder and wirtes it file name to db

// change below is your assigned mySQL username
$user = "root";

// change to the pw below is your assigned mySQL password
$pw = "";

// change to the database you have permission to connect to
$db = "method_music";



//mysql_connect("localhost", "root", ""); 
$mysql_access = mysql_connect("localhost", $user, $pw);
mysql_select_db($db, $mysql_access);





$query  = "select distinct artist from songs order by artist";
$result = mysql_query($query);

//echo $query;
//exit(); 


?>

<html>

<head>
<meta http-equiv="Content-Type"
content="text/html; charset=iso-8859-1">
<title>Untitled </title>
</head>

<body>

<br>

<br>

<br>

<br>
<br>
<table border="1" cellspacing="1" width="100%"
bordercolorlight="#008ED2" id="table1"
style="BORDER-COLLAPSE: collapse">
    
    <tr>
        <td colspan="3" bgcolor="#323E8D"><p align="center"><font
        color="#FFFFFF"><b>:.Singers ..:</b></font></p>
        </td>
    </tr>
  <tr>
        <td valign="top"><table border="0" cellpadding="2">
            
        
  My code. The red line shows the line 72:
<?
//this line for dropdownbox
$option = "";
$numbers=0;
while($row = mysql_fetch_assoc($result)) ==> error point
{
$numbers++; // counting number of records
//This 2 line for drop downbox
extract ($row);
    $option .= "<option value=\"$artist\">$artist</option>\r\n";

    //echo "<br>Id : {$row['id']} <br>" .
     //    "filename :{$row['filename']} <br>" .
    //     "artist : {$row['artist']} <br>" . 
    //     "album : {$row['album']} <br>" .
    //     "title : {$row['title']} <br><br>";
 

?>


           
<tr>
                <td style="BORDER-BOTTOM-STYLE: dotted"><a
                href="albums.php?albumname=<?=strval( $row['artist'] );?>"><? echo $numbers; ?>=><?=strval( $row['artist'] );?></a> [<a
                href="javascript:newWindow('RateSingers.php?SingerID=1378')">Rate
                Me</a>] Hits:<b>???</b></td>
            </tr>

 


<?


} 


?>

**noielen** · Apr 27th, 2006, 09:05 PM

Try this in your query:

Code:

$query  = "select distinct `artist` from `songs` order by `artist`";

**tony007** · Apr 27th, 2006, 09:16 PM

Originally Posted by noielen

Try this in your query:

Code:

$query  = "select distinct `artist` from `songs` order by `artist`";

Thank u for u reply. Unfortuently same error !! Is it connection problem ?

**penagate** · Apr 27th, 2006, 09:22 PM

You must have no data in your database.

**noielen** · Apr 27th, 2006, 09:29 PM

what about:

Code:

$query = "select distinct(artist)... order by artist asc"

**tony007** · Apr 27th, 2006, 09:32 PM

Originally Posted by penagate

You must have no data in your database.

i check mysql db using phpmyadmin and i see there are data!

**tony007** · Apr 27th, 2006, 09:43 PM

Originally Posted by noielen

what about:

Code:

$query = "select distinct(artist)... order by artist asc"

Thank u for u reply.same error!

**noielen** · Apr 27th, 2006, 09:52 PM

i have this error before, i just put `` before and after of my fieldnames and tables. now i used that all in my queries...

**penagate** · Apr 27th, 2006, 09:52 PM

after

PHP Code:


$result = mysql_query($query);

add

PHP Code:


if (!is_resource($result)) echo mysql_error();

**tony007** · Apr 27th, 2006, 10:15 PM

Originally Posted by penagate

after

PHP Code:


$result = mysql_query($query);

add

PHP Code:


if (!is_resource($result)) echo mysql_error();

Thank u for u reply. I get this error:

No database selected

I do not know what i am doing wrong!!

**noielen** · Apr 27th, 2006, 10:19 PM

and thats....... obviously in your connection.

**noielen** · Apr 27th, 2006, 10:20 PM

try this

Code:

$link = mysql_select_db($db, $mysql_access);
$query  = "select distinct artist from songs order by artist";
$result = mysql_query($query, $link);

i'm not sure with this just try it.

**penagate** · Apr 27th, 2006, 11:17 PM

I think I know what it is, on a hosted machine the DB name is usually prefixed with your username, e.g.

PHP Code:


$db = 'tony007_method_music';

**noielen** · Apr 28th, 2006, 12:05 AM

Originally Posted by penagate

I think I know what it is, nn a hosted machine the DB name is usually prefixed with your username, e.g.

PHP Code:


$db = 'tony007_method_music';

mybe

**visualAd** · Apr 28th, 2006, 01:03 AM

You should be error checking every call to mysql which could fail. mysql_select_db() is the same as calling mysql_query('use dbname').

Code:

if (! mysql_select_db($db)) {
    echo(mysql_error());
    die;
}

if (! ($result = mysql_query($query)) {
    echo(mysql_error());
    die;
}

**penagate** · Apr 28th, 2006, 05:06 AM

use dbname

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