【发布时间】:2017-10-05 16:28:18
【问题描述】:
我正在尝试内部连接来自 OS TICKET 数据库的 3 个表。
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";
代码正在返回:
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 = '.patrick.kershner@gmail.com.') ORDER BY qbcd_ticket.ticket_id DESC"
但它在 while 子句中没有显示任何内容:
while ($row = mysqli_fetch_assoc($result)){
echo $row['qbcd_ticket.number]."<br>";}
我不确定发生了什么,或者为什么它不显示结果。
有人可以查看我上面的代码并进行验证吗?
【问题讨论】:
-
你对
$result变量做了什么尝试? -
$qry = "SELECT qbcd_user_email.address, qbcd_user_email.user_id, qbcd_ticket.number, qbcd_ticket.id FROM qbcd_user_email INNER JOIN qbcd_user ON qbcd_user.id = qbcd_user_email.user_id INNER JOIN qbcd_ticket ON qbcd_ticket.user_id WHERE (qbcd_user_email.address = '$email') ORDER BY qbcd_ticket.ticket_id DESC"; $result = mysqli_query($link, $qry); var_dump($qry)."<br>"; while ($row = mysqli_fetch_assoc($result)){ echo $row['qbcd_ticket.number']."<br>"; }
标签: php mysqli inner-join