我想内部连接3表是从OS TICKET数据库。不显示结果与内部加入3表
我使用的代码是$qry = "SELECT qbcd_user_email.address, qbcd_user_email.user_id FROM qbcd_user_email INNER JOIN qbcd_user ON qbcd_user.id = qbcd_user_email.user_id INNER JOIN qbcd_ticket ON qbcd_ticket.user_id WHERE (qbcd_user_email.address = '.$email.') ORDER BY qbcd_ticket.ticket_id DESC";
代码将返回:
string(287) "SELECT qbcd_user_email.address, qbcd_user_email.user_id FROM qbcd_user_email INNER JOIN qbcd_user ON qbcd_user.id = qbcd_user_email.user_id INNER JOIN qbcd_ticket ON qbcd_ticket.user_id WHERE (qbcd_user_email.address = '[email protected]') ORDER BY qbcd_ticket.ticket_id DESC"
,但它不是,而子句中任何显示:
while ($row = mysqli_fetch_assoc($result)){
echo $row['qbcd_ticket.number]."<br>";}
我不知道是什么继续,或为什么不显示结果。
有人可以检查我的代码,并验证?
你用'$ result'变量试过了什么? –
'$ QRY =“SELECT qbcd_user_email.address,qbcd_user_email.user_id,qbcd_ticket.number,qbcd_ticket.id FROM qbcd_user_email INNER JOIN qbcd_user ON qbcd_user.id = qbcd_user_email.user_id INNER JOIN qbcd_ticket ON qbcd_ticket.user_id WHERE(qbcd_user_email.address =“$电子邮件')ORDER BY qbcd_ticket.ticket_id DESC“; $ result = mysqli_query($ link,$ qry); var_dump($ qry)。“
”; while($ row = mysqli_fetch_assoc($ result)){ \t echo $ row ['qbcd_ticket.number']。“
”; \t }' – PKershner