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内容摘要
帝国cms调用会员名及投稿数量排名
文章正文
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]
代码注释
- 我的微信
- 这是我的微信扫一扫
- 我的微信公众号
- 我的微信公众号扫一扫