这里我使用的是JOIN查询,它将超级有效,合并两个表值后我只返回一个表,我不知道如何返回两个表的值。从这段代码中返回了task表的值,task_employe表的值我无法返回,我不知道该怎么办??
$dapartment = $_POST['department'];
$q = mysql_query("SELECT * FROM task_employee te, task t WHERE te.emp_designation='$dapartment' AND te.emp_id = t.t_assign_to");
$data = array();
while($r = mysql_fetch_assoc($q)){
$data[] = $r;
}
$count = sizeof($data);
if($count > 0){
$return=array('status'=>'success','count'=>sizeof($data),'data'=>$data);
echo json_encode($return);
}else{
$return=array('status'=>'not-found','count'=>sizeof($data),'data'=>$data);
echo json_encode($return);
}
内部联接的标准方式是:
SELECT * FROM task_employee te JOIN task t ON te.emp_id = t.t_assign_to WHERE te.emp_designation='$dapartment'