Cita:
este es el codWarning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/rymchile/public_html/manager/a_/Admin/principal.php on line 65
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/rymchile/public_html/manager/a_/Admin/principal.php on line 71
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/rymchile/public_html/manager/a_/Admin/principal.php on line 65
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/rymchile/public_html/manager/a_/Admin/principal.php on line 71
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/rymchile/public_html/manager/a_/Admin/principal.php on line 71
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/rymchile/public_html/manager/a_/Admin/principal.php on line 65
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/rymchile/public_html/manager/a_/Admin/principal.php on line 71
Código PHP:
//status...
$status = $row2["ID_STATUS"];
$sql4="Select NOMBRE_S From STATUS WHERE ID_STATUS=$status";
$resultado3=conexion1($sql4);
$cont=mysql_fetch_array($resultado3);
//proyecto...
$proyecto = $row2["ID_PROYECTO"];
$sql5="Select NOMBRE_P From PROYECTO WHERE ID_PROYECTO=$proyecto";
$resultado2=conexion1($sql5);
$cont2=mysql_fetch_array($resultado2);
printf("<tr><td> %s</td><td> %s </td>
<td> %s</td>
<td><a href=\"../tarea/detalleTar.php?id=%d&idEmp=%d\">Detalle</a></td></tr>",
$row2["NOMBRE_T"],$cont["NOMBRE_S"],$cont2["NOMBRE_P"],
$row2["ID_TAREA"],$idEmp);
$nomPro=$cont2["NOMBRE_P"];
}
?>
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