tengo esta consulta
sprintf("SELECT * FROM calificar_tarea WHERE id = %s and rfcalumno=%s", GetSQLValueString($colname_id, "int"), GetSQLValueString($colname_alumno, "text"));
$calificar = mysql_query($query_avisos, $connection) or die(mysql_error());
$row_calificar = mysql_fetch_assoc($calificar);
y esto para cambiar la imagen
<?
mysql_select_db($database_connection, $connection);
$res = mysql_query($avisos) or die(mysql_error());
if ($num = mysql_fetch_row($res)) {
if ($num[0] == 0) {
echo '<img src="../images/Tarea.png" width="25" height="25" />';
} else {
echo '<img src="../images/TareaRevisada.png" width="25" height="25"/>';
} }
solo q tengo un error You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'Resource id #8' at line 1
seguire intentando