我正在连接这两个表,但给出的错误是mysql_fetch_array()期望参数1为资源,$result=mysql_query("SELECT * FROM `photo_gallery`.`photographs` WHERE id=1");while($row=mysql_fetch_array($resul
invoice_hearing_aids t2 WHERE t2.contact_id = invoice_hearing_aids.contact_id) ) sel
LEFT JOIN hearing_aid_makes makes ON makes.id= sel.make_id
LEFT JOIN contacts ON contacts.id = sel.
这个想法是,在给定的一天中有两种用餐选择,例如:10/02/16 meal id 4 OR meal id 12select * from meals JOIN menu ON meals.id = menu.option1 AND meals.id=menu.option2
但每次我运行这个命令时,我都会收到一条错误消息