2014-10-10 55 views
-1

我有2個表
表1結構:日誌]
用戶ID ||點擊||日期|| XXX || XXX |||
表2結構:[用戶]
username ||電子郵件||日期|| XXX || XXX |||
如何收集從使用兩個表中的數據連接

我想這樣

用戶名顯示的數據,COUNT(點擊)

我使用下面的查詢。

SELECT `user.username`,`userid`, COUNT(`userid`) as `total` 
FROM `logs` 
    INNER JOIN user ON logs.userid=user.userid 
ORDER BY total DESC 
LIMIT 5 

請quide

第一答覆後,我體改查詢和應用。查閱全文,就像這樣

$sqlptcwall=mysql_query("SELECT user.username, COUNT(ptcwalllogs.userid) AS total FROM logs INNER JOIN user ON user.userid = ptcwalllogs.userid GROUP BY user.username ORDER BY COUNT(ptcwalllogs.userid) DESC 
LIMIT 5"); 



<?php 
echo "</tr>"; 
     while($row = mysql_fetch_array($sqlptcwall)) 
     { 
echo "<tr> "; 
    echo "<td>" .$row[userid] . "</td>"; 
    echo "<td>" .$row[total] . "</td>"; 
    } 
echo "</tr> " ; 
?> 

錯誤: mysql_fetch_array()預計參數1是資源,布爾給定在

+0

只需添加一個'group by [some_field]' – Barranka 2014-10-10 18:02:27

回答

1
SELECT user.username, COUNT(user.userid) AS total 
FROM logs 
    INNER JOIN user ON user.userid = logs.userid 
GROUP BY user.username 
ORDER BY COUNT(user.userid) DESC 
LIMIT 5 
+0

edit t他質疑我收到的更多細節和錯誤 – nasima 2014-10-10 18:15:40

+0

您現在的問題不在原始問題的範圍之內。仔細閱讀錯誤消息並檢查您的數據訪問代碼。 – Namrehs 2014-10-10 18:17:40

+0

發現我的錯誤:) – nasima 2014-10-10 18:22:58