Question

我有2个表用户|讯息。

posts结构是：

[post_id | post_user_id]

users结构是：

[user_id | user_name | user_password]

现在我希望获得帖子的前5位用户

编辑1：

       $cod_sql = "select u.user_name, count(*) as 'TotalNumberofPosts'
from users u
  join posts p
    on u.user_id=p.post_user_id
group by user_name
order by count(*) desc 
limit 5";
        $top10 = mysql_query($cod_sql1) or die (error_sql(mysql_error(),__LINE__,__FILE__));

            echo '<div class="head_panel">Members</div> 
                    <div class="body_panel">
                <table border="1" width="100%">
                    <tr>
                    </tr>   
                    <tr class="tr3">
                Top5 : 
                    </tr>';

    if ( $top10 === FALSE ) 
    { 
        // An error has occured 
        print "There was an error running the query" . mysql_error(); 
    } else { 
        while ( $row = mysql_fetch_assoc($top10) ) 
        { 
            print "This is some data from a row<BR />"; 
            print "Name: ($row) <BR />"; 
        } 
    }  

          }
        break;

Answer 1

我想这会做。我现在无法验证脚本，所以请告诉我它是否有效。

select * 
from users
where exists (
     select * from posts
     where users.user_id=posts.post_user_id)
order by users.user_id
limit 5

Answer 2

以下代码将计算每个用户的帖子总数，并仅显示前5个海报。 SQL Fiddle demonstration

select u.user_name, count(*) as 'TotalNumberofPosts'
from users u
  join posts p
    on u.user_id=p.post_user_id
group by user_name
order by count(*) desc 
limit 5

有帖子mysql的热门用户

2 个答案: