Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource
Code (PHP)
<?php
$sql_dis="Select * from tb_members a
inner join tb_project_reports b on (a.tb_member_id = b.tb_member_id)
inner join tb_person_data c on (a.tb_member_id = c.tb_member_id)
inner join tb_work_today d on (a.tb_member_id = d.tb_member_id)
inner join tb_work_tomorrow e on (a.tb_member_id = e.tb_member_id)
where a.tb_member_id ='".$tb_member_id."' order by pjt_id desc";
$query_dis=mysql_query($sql_dis);
$fetch_dis=mysql_fetch_array($query_dis);
?>
แต่ถ้า join สองตารางได้ครับ
Code (PHP)
<?php
$sql_dis="Select * from tb_members a
inner join tb_project_reports b on (a.tb_member_id = b.tb_member_id)
where a.tb_member_id ='".$tb_member_id."' order by pjt_id desc";
?>