Simple SQL Select Nightmare

[darthmahon]

Right, SQL select problem really annoying me as it's a simple query.

This is the address:

meetings.php?case=2&area=Bristol

Now this is the code to select the data.

if($case==2)
{
$res_dl = mysql_query ("
SELECT * FROM MEETINGS WHERE area=$area");
$num_dl = mysql_num_rows ($res_dl);
}

What happens then is I get the following php reply at the top:

Warning: Supplied argument is not a valid MySQL result resource in /home/public_html/members/meetings.php on line 67 

Line 67 is:

$num_dl = mysql_num_rows ($res_dl);

Any ideas? This is such a simple query I don't understand the problem.

Cheers

[okuto1973]

Originally posted by darthmahon
Right, SQL select problem really annoying me as it's a simple query.

This is the address:

meetings.php?case=2&area=Bristol

Now this is the code to select the data.

if($case==2)
{
$res_dl = mysql_query ("
SELECT * FROM MEETINGS WHERE area=$area");
$num_dl = mysql_num_rows ($res_dl);
}

What happens then is I get the following php reply at the top:

Warning: Supplied argument is not a valid MySQL result resource in /home/public_html/members/meetings.php on line 67 

Line 67 is:

$num_dl = mysql_num_rows ($res_dl);

Any ideas? This is such a simple query I don't understand the problem.

Cheers [/B]

Have you created a connection to the DB?

<?
$db_name='dbname';
$db_user='user';
$db_password='password';

// connessione al DATABASE
$db_conn = mysql_connect("localhost",$db_user,$db_password);
        if (!$db_conn)
                {
                echo "ERRORE !! non riesco a collegarmi al database!";
                exit();
		}
?>

and

if($case==2)
{
$select="SELECT * FROM MEETINGS WHERE area=$area";
echo "select=$select";
$res_dl =mysql_db_query($db_name,$select);
$num_dl = mysql_num_rows ($res_dl);
echo "num_dl=num_dl";
}

So you can also check if you pass the correct string to the db......

[darthmahon]

Got a connection to the database set up yep.

Put in that echo select as you suggested and it returned the following:

select=SELECT * FROM MEETINGS WHERE area=Brighton Warning: Supplied argument is not a valid MySQL result resource in /home/public_html/members/meetings.php on line 68 num_dl=num_dl 

Thing is, I have another query like this:

if ($case==1)
{
$res_dl = mysql_query ("
SELECT * FROM MEETINGS WHERE areaid=$areaid ORDER BY morder");
$num_dl = mysql_num_rows ($res_dl);
}

Now, the variable $areaid is only numbers, not text invovled. So the address would look like:

meetings.php?case=1&areaid=7

That works fine and is on the same page so?? Bah! =/

[Bijan]

Try this one:

if($case==2) 
{ 
$res_dl = mysql_query ("SELECT * FROM MEETINGS WHERE area='$area'"); 
$num_dl = mysql_num_rows ($res_dl); 
}

Did you notice those extra '' marks that I put around $area? You should always do it for your string variables. Also it's a good idea that b4 you use any function on your mysql returned result set, you see if your query was excecuted successfully, you can test it like:

if($case==2) 
{ 
$res_dl = mysql_query ("SELECT * FROM MEETINGS WHERE area='$area'"); 
if ( !$res )
    die("Couldn't do the query successfully, because " mysql_error() );
$num_dl = mysql_num_rows ($res_dl); 
}

[okuto1973]

The correct one:

 
if($case==2)  

{  

$res_dl = mysql_query ("SELECT * FROM MEETINGS WHERE area='$area'");  

if ( !$res_dl ) 
    die("Couldn't do the query successfully, because " mysql_error() ); 
$num_dl = mysql_num_rows ($res_dl);  

}

[Bijan]

Oopse, thanks for your correction, I really didn't notice the variable name!

[darthmahon]

thanx bijan, works a treat =)