MYSQL ERROR not a valid MySQL result resource

[spighita]

Hello:
I have a php script where on localhost works fine, but when I upload it to the server I get this error

Warning: Supplied argument is not a valid MySQL result resource

The script is

<TABLE border=1>
 <TR> <TD><STRONG>No </STRONG></TD> <TD><STRONG>Yes</STRONG></TD></TR>
 <TR>  

 <?PHP
	$con = mysql_connect("IP", "", "") or die("");
	mysql_select_db("database", $con);
	$query = "SELECT COUNT(*) FROM DATABASE GROUP BY groupnum";
	$num= mysql_query($query, $con);
	while ($rs = mysql_fetch_array($num)){
		echo '<TD align="center">', $rs[0]; '</TD>';

	}


?>



</TR>
 </TABLE>

thanks in advance for anyone who helps

[stolzyboy]

use this

$num= mysql_query($query, $con) or die(mysql_error());

instead of this

$num= mysql_query($query, $con);

this will tell you why it isn't valid, most likely your query isn't returning anything

[spighita]

I used it and got no error or errno. The line problem appears when this is read

while ($rs = mysql_fetch_array($num)){

that`s where I get the error

This works perfect on my computer localhost but not when uploaded to the server

[barand]

I notice that you select a database called 'DATABASE' and you also query a table called 'DATABASE'.

Do you have a table in database 'database' called 'database' ?

[spighita]

Yes you are right, I changed them quickly cause I had the real IP and the database name, the select count(*) should be plural.

Its real name is databases, just an example for the name

It does select the database. It work perfect on my computer. but when I upload it that error appears.

I have made some research on the web and found that I'm not the only one with the problem.

look at this link

http://forums.devshed.com/archive/4/2000/06/1/2310

[TheIceman5]

geez "or die(""); " thats an imformative error message, top points for that.
And wheres your username and password info?? dont see it.

[spighita]

Its not necessary. There is no point inserting it cause my issue is the error Im getting.

There is no difference whether what error message will appear. The whole script is not in english unless you would like to learn some italian or spanish error string, i`d post them gladly.

[barand]

The things likely to change between home and ISP are servername, user and password. Are you sure you are connecting OK?

[spighita]

Yes, the script is big, and the only problem is that one. That only line.

[barand]

The error isn't actually on that line, it's happening higher up. Either the connect or the query itself. It's just that it does not show up until you try to use the (invalid) query result.

[spighita]

But if I remove that that mysql query everything works OK.

The php build that the host has is 4.1.2 and have the latest. Can that make a difference?

[barand]

To GROUP BY groupnum,

SELECT groupnum, count(*) FROM ...

[cgraz]

I believe the result resource the error is describing is $con. $query looks to be ok (assuming all your names etc are correct and properly written). As stolzyboy suggested, use mysql_error when executing your query, but you may also want to give it a try when using mysql_connect just to make sure mysql is establishing a connection. You would never know if it wasn't because the error it is printing is "". This is my guess to the problem. It "works" when you remove the query because you don't see the error message produced by that query. But if I'm right, then it's not actually "working" without the query, you just don't know there's a problem because of your error-handling. "" prints nothing. Try this instead

$con = mysql_connect("IP", "", "") or die(mysql_error()); 

Cgraz

[spighita]

OK, this is the script I'm using, but I still get the error

Warning: Supplied argument is not a valid MySQL result resource in /home/httpd/html/clientes/www.mysite.com/adm.php on line 75

<TABLE border=1>
 <TR> <TD><STRONG>No autorizados</STRONG></TD> <TD><STRONG>Autorizados</STRONG></TD></TR>
 <TR>  

 <?PHP
	$con = mysql_connect("localhost", "$user", "$pass") or die("No se pudo conectar con el servidor");
	mysql_select_db("socios", $con) or die (mysql_error());
	$query = "SELECT groupnum, COUNT(*) FROM SOCIO GROUP BY groupnum";
	$num= mysql_query($query, $con);
	while ($rs = mysql_fetch_array($num)){
		echo '<TD align="center">', $rs[0]; '</TD>';

	}


?>



</TR>
 </TABLE>

[barand]

If its getting past the connection then must be the query itself. Check error message there too. Perhaps "SOCIO" should be "socio".

<?PHP 
    $con = mysql_connect("localhost", "$user", "$pass") or die("No se pudo conectar con el servidor"); 
    mysql_select_db("socios", $con) or die (mysql_error()); 
    $query = "SELECT groupnum, COUNT(*) FROM SOCIO GROUP BY groupnum"; 
    $num= mysql_query($query, $con) or die(mysql_error()); 
    while ($rs = mysql_fetch_array($num)){ 
        echo '<TD align="center">', $rs[0]; '</TD>'; 

    } 


?> 

[spighita]

YESSSSSSSSSSSSSSSSSSs,

YOU FIXED IT. Also I had to change

SELECT groupnum, COUNT(*) FROM SOCIO GROUP BY groupnum

to SELECT COUNT(*) FROM socio GROUP BY groupnum

Thank everyone.

Thank you BARAND.