1、調(diào)用會(huì)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(huì)員發(fā)布文章數(shù)，增加（序號(hào)、會(huì)員id）

<table><tr><td>排名號(hào)</td><td>會(huì)員名</td><td>文章數(shù)</td><td>會(huì)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(huì)出項(xiàng)一個(gè)問題就是 管理員的ID會(huì)與前臺(tái)會(huì)員的ID重復(fù)

即：管理員的ID=1，前臺(tái)會(huì)員的ID=1（所以管理員的ID=前臺(tái)會(huì)員的ID），最后統(tǒng)計(jì)出來的文章會(huì)是：管理員+前臺(tái)會(huì)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(hào)</td><td>會(huì)員名</td><td>文章數(shù)</td><td>會(huì)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>