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>

发表评论

后才能评论