mysql_num_rows(): error

[phphelp_]

Hello I am very new at php, and needed some help.

I get an error of:
"Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in /usr/home/www.xxx.com/xxx/xxx/online.php on line 176"
The code is:

$content.= $s;
   }
}
function showAll(){
   global $PATHTO_IMAGES, $DB_ONLINEPUB, $content;

   $mysqlDataAllPub   = mysql_query( "SELECT * FROM $DB_ONLINEPUB  WHERE  `publish` =  '1' ORDER BY `name` ASC"); 
   $dataLengthAllPub = mysql_num_rows($mysqlDataAllPub); //echo "HELLO DATALENGTH=" .$dataLengthAllPub;  

   $count = 0;
   $s='';
      for ($i=0; $i< $dataLengthAllPub; $i++ ) {
      $online_pubID     = mysql_result( $mysqlDataAllPub , $i, pubID    );
      $online_name      = mysql_result( $mysqlDataAllPub , $i, name     );
      $online_orderID   = mysql_result( $mysqlDataAllPub , $i, orderID );
      $s.= '<li >
           <div id="showallmod">
              <div class="con">'
              .showAllIssues($online_pubID).
           '</div>
           </div>
           </li>';
      }   

   $content= $s;
   }

it looks like my query is not being populated properly. I checked my query format and got:
You have an error in your SQL syntax. Check the manual that corresponds to your MySQL server version for the right syntax to use near 'WHERE publish = '1' ORDER BY name ASC' at line 1
QUERY: SELECT * FROM WHERE publish = '1' ORDER BY name ASC
My variable $DB_ONLINE is not being passed to my showall() function.

I believe that where the variable defined is:

if($pubID=="" && $issueID=="" || $task == "showall"){ 
   $task="showall";
   showAll();
   }
if($task=="" && $issueID){
   fetchIssueItem(); 
   fetchPages($issueID);
}
if($task=="fetchissues"){
   fetchPubItem(); 
   }
if($task=="detail"){
   fetchIssueItem(); 
   fetchPages($issueID);
}
function displayListPubMenu(){
   global $PATHTO_IMAGES, $DB_ONLINEPUB;
   $mysqlDataListPub   = mysql_query( "SELECT * FROM $DB_ONLINEPUB  WHERE  `publish` =  '1' ORDER BY `name` ASC"); //SELECT * FROM $DB_WALLPAPER where id!=$id ORDER BY $order_by" );
   $dataLengthPub = mysql_num_rows($mysqlDataListPub);

   $s='';
      for ($i=0; $i< $dataLengthPub; $i++ ) {
      $online_pubID    = mysql_result( $mysqlDataListPub, $i, pubID    );
      $online_name      = mysql_result( $mysqlDataListPub, $i, name     );
      $online_orderID   = mysql_result( $mysqlDataListPub, $i, orderID );

   $s.= '<li><a href="online.php?task=fetchissues&pubID='.$online_pubID.'" target="_self">'. $online_name . '</a></li>';   

   }   

   return $s;

This is the point I am stuck at any help would be greatly appreciated!

[dagon]

it means the query that is returning the rows your trying to count failed. try adding some error checking

example:

$mysqlDataListPub = mysql_query( "SELECT * FROM $DB_ONLINEPUB WHERE `publish` = '1' ORDER BY `name` ASC"); //SELECT * FROM $DB_WALLPAPER where id!=$id ORDER BY $order_by" );
if (!$mysqlDataListPub) {
    die('Invalid query: ' . mysql_error());
}
$dataLengthPub = mysql_num_rows($mysqlDataListPub);

[phphelp_]

Hey dagon,
Thank you for the reply.
I get a message of:
Invalid query: You have an error in your SQL syntax. Check the manual that corresponds to your MySQL server version for the right syntax to use near 'WHERE publish = '1' ORDER BY name ASC' at line 1

[bradgrafelman]

Sounds like $DB_ONLINEPUB is undefined or empty.

Whenever you have SQL query errors, it often helps to echo out the full query string as well. It is for this reason that I normally follow this format when writing SQL queries:

$sql = 'SELECT * FROM foo'; // put the actual query in a variable

// remove the die(), of course, on production server:
$exec = mysql_query($sql) or die("MySQL error: " . mysql_error() . "<br>\nQuery: $sql");

[tommyrulez]

I maybe wrong, but does he need the '' around the 1?

WHERE `publish` = '1'

[bradgrafelman]

tommyrulez;10945632 wrote:

I maybe wrong, but does he need the '' around the 1?

WHERE `publish` = '1'

If the 'publish' column is a numeric type, then no, there shouldn't be quotes around its value. That's not, however, causing the present SQL error.

[phphelp_]

Hello,
Thank you so much for all the help, as I said I am very new at php.
Where do I put the code you just gave me? do I put it on top of the line I have an error on, do I replace it?
I am working off someone else's code so I really have no idea what all this means. I am trying to learn my way through this battle field called php.
So any help would be great, and again I need to be told exactly what and where to do it. Thank you all again