php - 如何以更简洁的方式执行这些 MySQL 子查询？

Question

使用使用mysql而不是mysqli/pdo的遗留代码，所以不用担心这个，我稍后会更新这个查询。

尽管我当前的方法有效，但我确信有一种更简洁的方法可以完成此操作，而不是使用查询和 3 个子查询。我主要想了解如何更好地增强查询并最大程度地减少查询量。
我想做的是

1. 回显每个日期的所有数据，日期显示在顶部
2. 在用户旁边显示该特定日期每个用户的条目数

3. 对于每个日期，在上述 2 位数据的底部，显示条目数最多的用户

   $query = mysql_query('SELECT * FROM entries GROUP BY DATE(dt)');
   
   $g = 0;
   
   while ($row = @mysql_fetch_array($query)) 
   {
   
       $group[$g] = date('y-m-d', strtotime($row['dt']));
       echo $group[$g] . "<br />";
   
       //display the person's name for today with their count
       $dayquery = mysql_query('SELECT *, COUNT(username) as total FROM entries WHERE DATE(dt) = "'.$group[$g].'" GROUP BY username ORDER BY COUNT(username) DESC');
           while ($today = @mysql_fetch_array($dayquery)) 
           {
               echo $today['first_name'] . " | " . $today['total'] . "<br />";
           }   
   
           //display the highest count for today
           $topquery = mysql_query('SELECT COUNT(username) as highest FROM entries WHERE DATE(dt) = "'.$group[$g].'" GROUP BY username ORDER BY COUNT(username) DESC LIMIT 1');
               while ($toptoday = @mysql_fetch_array($topquery)) 
               {
                   echo "Highest today: " . $toptoday['highest'] . "<br /><br />" ;
               } 
   
               //display the users with the highest count for today
               echo "Highest users: ";
               $userstopquery = mysql_query('SELECT *, COUNT(username) as total FROM entries WHERE DATE(dt) = "'.$group[$g].'" AND COUNT(username) = "' . $toptoday['highest'] . '" AND GROUP BY username');
                   while ($topusers = @mysql_fetch_array($userstopquery)) 
                   {
                        echo $topusers['first_name'] . "<br />" ;
                   } 
   
       $g++;
   
   }
   

我遇到的麻烦是，当我尝试减少这些子查询并使用 MAX 时，它只会输出最高计数，而不是每个日期的所有数据，这正是我所需要的，包括当天条目数量最多的用户的输出。

Answer 1

你可以从这样的事情开始。请注意，我没有使用 PHP mysql API，因为它在 3 或 4 年前已被弃用...

require('path/to/mysqli/connection/stateme.nts');

$array = array();

$query = "
SELECT e.dt
     , e.username
     , COUNT(*) ttl
  FROM entries e
 GROUP
    BY e.dt
     , e.username
 ORDER 
    BY e.dt, ttl DESC;
";


$result = mysqli_query($conn,$query);

while ($row = mysqli_fetch_assoc($result)) 
        {
            $array[] = $row;
        }

print_r($array);