Please help me with this php code - getting errors

I keep getting this error:

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource on first line of this code:

The code is:

  $sql = "select * from sponsored_jokes where jokeid = $jokeid";

   $result = mysql_query($sql ,$db);

   if ($myrow = mysql_fetch_array($result)) {

     do {

        $sponsorid = $myrow["sponsorid"];

     } while ($myrow = mysql_fetch_array($result));

   }

Code (markup):

It worked fine on PHP4. Now it does not on PHP5.

Please help. Thank you.

 

Add this after the mysql_query() line: (It should show us the error message)

if (!$result) {
    die('Invalid query: ' . mysql_error());
}

PHP:

 

Thank you. I get these:

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 '' at line 1

 

I think the syntax is correct but it looks better like this:

$sql = "SELECT * FROM sponsored_jokes WHERE jokeid = $jokeid";

PHP:

Most importantly, Is $jokeid null?

 

Could you also add echo $sql;

 

I think the script can't see the $jokeid value...... but it seems to connect to the database!

 

Well, where are you getting $jokeid from? Where is it initialized?

Shouldn't it be $_GET['jokeid']? (be sure to use mysql_real_escape_string($_GET['jokeid']) in your SQL)

 

even when I write $sql = "select * from sponsored_jokes where jokeid = 2";

i still get that error instead of displaying joke number 2

 

Try to do that same SQL in phpmyadmin and see what it says there.

 

Try:

  $sql = "SELECT * FROM `sponsored_jokes` WHERE `jokeid` = '$jokeid'";

   $result = mysql_query($sql, $db) or die(mysql_error());

   if ($myrow = mysql_fetch_array($result)) {

     do {

        $sponsorid = $myrow["sponsorid"];

     } while ($myrow = mysql_fetch_array($result));

   }

Code (markup):

 

The error disappears but I get no content instead Thank you anyway.

 

$sponsorid is empty?

$sql = "SELECT * FROM `sponsored_jokes` WHERE `jokeid` = '$jokeid'";
$result = mysql_query($sql, $db) or die(mysql_error());
while($myrow = mysql_fetch_array($result)){
    $sponsorid = $myrow["sponsorid"];
}

Code (markup):

Not liking the do thing.

 

Still not working.... thank you anyway! It seems the script can't access the db correctly.... however it does access it..but only partially.

 

add single quotes to var : $sql = "SELECT * FROM sponsored_jokes WHERE jokeid = '".$jokeid."'";

 

is $myrow set anywhere before this line:
if ($myrow = mysql_fetch_array($result))
?

 

thank you everyone. the problem was solved with a script upgrade.