1、調(diào)用會員發(fā)布文章數(shù)

<table>
[e:loop={'SELECT userid, username, count(username) as total from [!db.pre!]ecms_news group by username order by total desc',0,24,0}]
<tr>
<td><?=$bqno?></td>
<td><?=$bqr[username]?></td>
<td><?=$bqr[total]?></td>
</tr>
[/e:loop]
</table>

2、只調(diào)用會員發(fā)布文章數(shù)，增加（序號、會員id）

<table>
<tr>
<td>排名號</td>
<td>會員名</td>
<td>文章數(shù)</td>
<td>會員ID</td>
</tr>
[e:loop={'select userid, username,count(username) as num from [!db.pre!]ecms_news group by username order by num desc',0,24,0}]
<tr>
<td><?=$bqno?></td>
<td><?=$bqr[username]?></td>
<td><?=$bqr[num]?></td>
<td><?=$bqr[userid]?></td>
</tr>
[/e:loop]
</table>

注釋：在sql語句“ SELECT userid, username, count(username) as total from [!db.pre!]ecms_news group by username order by total desc ”

中的“(username)”和“group by username”中的 “username”也能用 “userid” 調(diào)用 但會出項一個問題就是 管理員的ID會與前臺會員的ID重復

即：管理員的ID=1，前臺會員的ID=1（所以管理員的ID=前臺會員的ID），最后統(tǒng)計出來的文章會是：管理員+前臺會員=總數(shù)

月排行

where newstime > UNIX_TIMESTAMP()-86400*30 （月：30、周：7）

舉例：月排行

<table><tr><td>排名號</td><td>會員名</td><td>文章數(shù)</td><td>會員ID</td></tr>
[e:loop={'select userid, username,count(username) as num from [!db.pre!]ecms_news where newstime > UNIX_TIMESTAMP()-86400*7 group by username order by num desc',0,24,0}]
<tr><td><?=$bqno?></td><td><?=$bqr[username]?></td><td><?=$bqr[num]?></td><td><?=$bqr[userid]?></td></tr>
[/e:loop]
</table>