![enojado](http://static.forosdelweb.com/fdwtheme/images/smilies/enojado.png)
Mi codigo:
Código PHP:
while($CONTAINER = mysql_fetch_array($containerq))
{
while($CHAR = mysql_fetch_array($charges))
{
if(${'chargepp' . $CHAR[0] . 'cont' . $CONTAINER[0]} == "on")
{
$idchar = mysql_query("SELECT id_charge, ch_type FROM `toctrack_charges` WHERE `ch_type` = '1' ORDER BY id_charge DESC LIMIT 1") or die("Invalid query: " . mysql_error());
$CHARGEID = mysql_fetch_array($idchar);
mysql_query("INSERT INTO `toctrack_chargein` (`id_charges`, `id_cat_charges`)
VALUES ('$CHARGEID[0]', '$CHAR[0]')") or die("Invalid query: " . mysql_error());
}
else if(${'chargesp' . $CHAR[0] . 'cont' . $CONTAINER[0]} == "on")
{
echo "asdfñjdsafñlñfjdfklñjdsa";
$idchar2 = mysql_query("SELECT id_charge, ch_type FROM `toctrack_charges` WHERE `ch_type` = '2' ORDER BY id_charge DESC LIMIT 1") or die("Invalid query: " . mysql_error());
$CHARGEID2 = mysql_fetch_array($idchar2);
mysql_query("INSERT INTO `toctrack_chargein` (`id_charges`, `id_cat_charges`)
VALUES ('$CHARGEID2[0]', '$CHAR[0]')") or die("Invalid query: " . mysql_error());
}
}
}
Saludos!