我想內部連接3表是從OS TICKET數據庫。不顯示結果與內部加入3表
我使用的代碼是$qry = "SELECT qbcd_user_email.address, qbcd_user_email.user_id FROM qbcd_user_email INNER JOIN qbcd_user ON qbcd_user.id = qbcd_user_email.user_id INNER JOIN qbcd_ticket ON qbcd_ticket.user_id WHERE (qbcd_user_email.address = '.$email.') ORDER BY qbcd_ticket.ticket_id DESC";
代碼將返回:
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,而子句中任何顯示:
while ($row = mysqli_fetch_assoc($result)){
echo $row['qbcd_ticket.number]."<br>";}
我不知道是什麼繼續,或爲什麼不顯示結果。
有人可以檢查我的代碼,並驗證?
你用'$ result'變量試過了什麼? –
'$ 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 =「$電子郵件')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