[SOLVED] trouble with query

[jakebur01]

hey when I add the "WHERE catid != TBP " in I get this error: PHP Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, boolean given in

 

heres the code:

$result=mysqli_query($db, "SELECT isbn, description, catid, title, author ".
    "FROM books ".
"WHERE catid != TBP ".
"order by isbn asc ".
    "limit $offset,$limit"
);

 

any ideas??

[teng84]

$result=mysqli_query($db, "SELECT isbn, description, catid, title, author FROM books WHERE catid != TBP order by isbn asc limit $offset,$limit");

 

try that

[jakebur01]

ha.. I couldn't get that to work either. I even tried it like this:

 

 <?php


$db = mysql_connect("", "", "");
mysql_select_db("", $db);

$result = mysql_query("SELECT * FROM books WHERE catid <> TBP", $db);
while($myrow = mysql_fetch_array($result))
{
$catid = $myrow['catid'];
$title = $myrow['isbn'].' '.$myrow['description'].' '.$myrow['title'].' '.$myrow['author'];
$url = "show_cat.php?catid=$catid";
echo "<small";
?>
  <a href="<?php echo $url; ?>"><?php echo $title; ?></a><br />
<?php
echo"</small>";
echo "<br />";
}

 

I don't want any of the rows where TBP is catid. Basically... I've tried Where catid <> and i tried where catid !=  ...

 

Any suggestions?

 

[teng84]

whats the error message

 

[jakebur01]

PHP Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource

[teng84]

ok lets debug

remove the $db

and do you really have the connection any way try that first

$result = mysql_query("SELECT * FROM books WHERE catid <> TBP");

[teng84]

oops

$result = mysql_query("SELECT * FROM books WHERE catid <> 'TBP' ");

forgot put '' on TBP

[jakebur01]

oh man it worked!

 

Thank you.

[teng84]

;D