1、调用会员发布文章数
<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><?=$bqr[total]?></td>
</tr>
[/e:loop]
</table>2、只调用会员发布文章数，增加（序号、会员id）
<table>
<tr>
<td>排名号</td>
<td>会员名</td>
<td>文章数</td>
<td>会员ID</td>
</tr>
[e:loop={'select userid, username,count(username) as num from [!db.pre!]ecms_news group by username order by num desc',0,24,0}]
<tr>
<td><?=$bqno?></td>
<td><?=$bqr[username]?></td>
<td><?=$bqr[num]?></td>
<td><?=$bqr[userid]?></td>
</tr>
[/e:loop]
</table>
注释：在sql语句“ SELECT userid, username, count(username) as total from [!db.pre!]ecms_news group by username order by total desc ”
中的“(username)”和“group by username”中的 “username”也能用 “userid” 调用 但会出项一个问题就是 管理员的ID会与前台会员的ID重复
即：管理员的ID=1，前台会员的ID=1（所以管理员的ID=前台会员的ID），最后统计出来的文章会是：管理员+前台会员=总数
月排行
where newstime > UNIX_TIMESTAMP()-86400*30 （月：30、周：7）
举例：月排行
<table><tr><td>排名号</td><td>会员名</td><td>文章数</td><td>会员ID</td></tr>
[e:loop={'select userid, username,count(username) as num from [!db.pre!]ecms_news where newstime > UNIX_TIMESTAMP()-86400*7 group by username order by num desc',0,24,0}]
<tr><td><?=$bqno?></td><td><?=$bqr[username]?></td><td><?=$bqr[num]?></td><td><?=$bqr[userid]?></td></tr>
[/e:loop]
</table>
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