[SOLVED] Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result reso

[pontiac007]

I keep getting this:

 

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

 

code:

 

while($row = mysql_fetch_array($result, MYSQL_ASSOC))

 

what can i do? i still new in php.

[genericnumber1]

show us all your code

[papaface]

Should be:

 while($row = mysql_fetch_assoc($result))

[genericnumber1]

while($row = mysql_fetch_assoc($result))

 

is the same thing as

 

while($row = mysql_fetch_array($result, MYSQL_ASSOC))

[per1os]

We need to see the query and where $result is being set.

[pontiac007]

here is the codes:

 

$query = "SELECT acadstaff.idstaff, acadstaff.staffname, student.idstud FROM acadstaff, student WHERE idstud=$_SESSION[idstud]";

$result = mysql_query($query);

 

while($row = mysql_fetch_array($result, MYSQL_ASSOC))

{

$staffname= $row['staffname'];

?>

      <?php echo "{$row['staffname']} <br>";?></a>

        <?

}

 

Iive tried changing array to assoc before but it doesn't work.

[per1os]

<?php
$query = "SELECT acadstaff.idstaff, acadstaff.staffname, student.idstud FROM acadstaff, student WHERE idstud=$_SESSION[idstud]";
$result = mysql_query($query);

   while($row = mysql_fetch_array($result))
   {
   $staffname= $row['acadstaff.staffname']; 
echo $row['acadstaff.staffname'];

   }
?>

 

Should work.

[pontiac007]

It doesn't work.

 

I've tried to change SELECT statement but it still didn't work out.

 

[per1os]

<?php
$query = "SELECT acadstaff.idstaff, acadstaff.staffname, student.idstud FROM acadstaff, student WHERE idstud=$_SESSION[idstud]";
$result = mysql_query($query) or DIE(mysql_error());

   while($row = mysql_fetch_array($result))
   {
   $staffname= $row['acadstaff.staffname']; 
echo $row['acadstaff.staffname'];

   }
?>

 

See if there is an error in the query.

[pontiac007]

Thanx guys!! i've found my mistake.