Conultar varias tablas php y mysql

[horusorion]

$query = "SELECT schedules.id, schedules.employee_id, schedules.class, employees.name FROM schedules, employees WHERE schedules.employee_id = employees.id" or die("Error in the consult.." . mysqli_error($link));
$execute = mysqli_query($link, $query);
$rows= mysqli_num_rows($execute);
if ($rows > 0){
while($row = mysqli_fetch_array($execute )) {
echo $row["schedules.id"];
echo $row["schedules.class"];
echo $row["employees.name"];
}
}else{
echo "No hay datos";
}

La sintaxis correcta si no me equivoco es :

Código PHP:

$query = "SELECT schedules.id, schedules.employee_id, schedules.class, employees.name FROM schedules, employees WHERE schedules.employee_id = employees.id" );
$execute = mysqli_query($link, $query)or die("Error in the consult.." . mysqli_error($link);
$rows = mysqli_fetch_assoc($execute );
$num_rows= mysqli_num_rows($execute);
if ($num_rows > 0){
        while($row = mysqli_fetch_assoc($execute )) {
               echo $row["schedules.id"];
           echo $row["schedules.class"];
           echo $row["employees.name"];
         }
}else{
    echo "No hay datos";
}
 

Espero haber podido ayudarte... tenias row como el que te decia cuantos registros habia, por eso te daba el error.... y lo puse un poco mas bonito... jjajaja