2017-05-07 57 views
0

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?

+0

你用'$ result'變量試過了什麼? –

+0

'$ 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

回答

0

嘗試將號碼添加到您選擇的特性

$qry = "SELECT qbcd_user_email.address, qbcd_user_email.user_id, qbcd_ticket.number FROM qbcd_user_email INNER JOIN qbcd_user ON qbcd_user.id = qbcd_user_email.user_id INNER JOIN qbcd_ticket ON qbcd_ticket.user_id WHERE (qbcd_user_email.address = '.$email.') ORDER BY qbcd_ticket.ticket_id DESC" 
+0

我試過了,更新了它,我認爲它在我的where循環中,或者在實際的qry中。我在下面的評論中添加表格。 – PKershner

0

第一個表是:

qbcd_ticket: 
rows: 

ticket_id | number | user_id | user_email_id | status_id | dept_id | and more... 
5  | 762086| 2  |  0 |  1| 1 |  


the next is qbcd_user_email 
rows: 
id | user_id | flags | address 
2 | 2 | 0 | [email protected] 

the last is: qbcd_user 
id | org_id | default_email_id | status | name    | created    | updated 
2 | 0 | 2   |  0 | Patrick Kershner | 2017-03-03 10:44:28 | 2017-03-03 10:44:28 

,我需要顯示的信息,與客戶相關聯的所有相應門票的地方=電子郵件地址。

唯一不會改變的靜態變量是$_SESSION['user_email'];,它通過登錄到成員區域進行記錄。