mysql_num_rows(): error please help

[phphelp!]

Hello,

I am very new to php so please bare with me.

 

I am getting 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;

}

 

 

I'm not sure even where to begin troubleshooting this.

I thank you in advance for your help

 

[schilly]

start with this:

$mysqlDataAllPub   = mysql_query( "SELECT * FROM $DB_ONLINEPUB  WHERE  `publish` =  '1' ORDER BY `name` ASC") or die(mysql_error()); 

 

[phphelp!]

thank you Schilly,

Like I said I am very new to this.

I am guessing that you want the message after I replaced the code.

it is:

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

[wildteen88]

Where is the variable $DB_ONLINEPUB defined. Looks like your query is not being populated properly. To check your query is formatted correctly change

$mysqlDataAllPub   = mysql_query( "SELECT * FROM $DB_ONLINEPUB  WHERE  `publish` =  '1' ORDER BY `name` ASC") or die(mysql_error()); 

to

$mysqlDataAllPubSQL = "SELECT * FROM $DB_ONLINEPUB  WHERE  `publish` =  '1' ORDER BY `name` ASC";
$mysqlDataAllPub   = mysql_query( $mysqlDataAllPubSQL ) or die(mysql_error() . '<br />QUERY: ' . $mysqlDataAllPubSQL ); 

[phphelp!]

I then get a message of:

 

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

[wildteen88]

Yeah, The variable $DB_ONLINEPUB is not being passed to your showAll() function. Where is this variable defined?

[phphelp!]

Hey Thank you much wildteen88

 

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;

[phphelp!]

Hello,

I was wondering if it looks correct, or really what should be changed.

Is my variable defined incorrectly?

Any help or guidance would be wonderful.

Thank you all