Problem displaying result in table

[dookie]

Hi, im trying to display some data in a table (php + mysql)
this is the code to create the table

function hacer_lista_materia()
{
include("opcion.php"); //this just have $numfilas = 10
$page=$_GET["pag"];
if (($page=="") or ($page==1))
{	$offset=0; }
else
{	$offset=($page-1)*$numfilas;}

[B]$result_count=mysql_query("SELECT count(*) as total FROM materia", $conexion);[/B]

[B]while ($count=mysql_fetch_array($result_count, MYSQL_ASSOC))[/B]
{$total= ceil($count["total"]/$numfilas);}

[B]$result=mysql_query("select codmateria, nommateria from materia order by nommateria LIMIT '$numfilas' OFFSET '$offset';", $conexion) or die('Consulta fallida');[/B]

echo ("<blockquote><table border \"1\">\n");
echo ('<font face="Geneva, Arial, Helvetica, sans-serif" size="1">');
echo ('<tr><td><div align="center"><b>Materia</td><td><div align="center"><b>Modificar</td><td><div align="center"><b>Eliminar</td></tr>');
while ($row=mysql_fetch_array($result, MYSQL_ASSOC))
{
echo '<tr><td><div align="left">'.$row["nommateria"].'</td>';
echo '<td><div align="center"><a href="mant_materia_mod.php?codmateria='.$row["codmateria"].'"><img src="../images/edit.gif" width="15" height="15" border="0" alt="Modificar"></a></td>';
echo '<td><div align="center"><a href="mant_materia_del.php?codmateria='.$row["codmateria"].'"><img src="../images/delete.gif" width="15" height="15" border="0" alt="Eliminar"></a></td></tr>';
};
echo ("</table></blockquote>");
If ($total>1)
{
echo("<center><font face='Geneva, Arial, Helvetica, sans-serif' size='2'>");
echo("Pág: <a href='mant_materia.php?pag=1'>1</a>");
for ($i = 2; $i <= $total; $i++)
	{
	if ($page==$i)
		echo (" - $i");
	else
	echo (" - <a href='mant_frase.php?pag=$i'>$i</a>");
	}
echo("</center>");
}
mysql_close($conexion);
} 

i have errors in those lines in bold:
Warning: mysql_query(): supplied argument is not a valid MySQL-Link resource in C:\web\SGC\funciones.php on line 12

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in C:\web\SGC\funciones.php on line 14

Warning: mysql_query(): supplied argument is not a valid MySQL-Link resource in C:\web\SGC\funciones.php on line 17
Consulta fallida

the idea is having something like this:

i dont know whats wronge in those lines, i used som similar code to connect to DB to login to a system, using sentences like $row=mysql_fetch_array($result, MYSQL_ASSOC) above..
any help please?

thanks

[Installer]

You haven't established a connection to the database (at least not in the function's scope), as the first error tells you. If you do have a connection outside the function, you need to pass its resource variable ($conexion) to the function, or establish one inside the function.

The same goes for your third error. The second error is just a consequence of the first.

[dookie]

thanks, i had it outside the function, and soing something very similar with postgres and php it worked.. so i didnt know i had to include the connection inside the function...
just 1 problem now:

this query drop me "consulta fallida"
$result=mysql_query("select codmateria, nommateria from materia order by nommateria LIMIT '$numfilas' OFFSET '$offset'", $conexion) or die('Consulta fallida');

if i remove the last part (LIMIT '$numfilas' OFFSET '$offset'") works fine... are LIMIT and OFFSET supported in mysql?

thanks again

[Installer]

I don't believe MySQL has an "OFFSET" clause. Maybe it's something you can replace with "WHERE"?

Here is the manual section for "SELECT".

[dookie]

thanks man, it has OFFSET clause, but with mysql it doesnt need those ' ' for the variables...
i better do all over instead of modify those postgres sentences...

thanks again 😉