如何以更清晰的方式執行這些MySQL子查詢？

Question

它採用mysql而不是mysqli/pdo使用舊代碼，所以不要擔心這個，我以後會更新該查詢。

雖然我目前的方法工作，我敢肯定有這樣做，而不是查詢和3子查詢的更清潔的方式。我主要想學習如何更好地增強我的查詢並儘量減少它們的數量。
我想要做的是

1. echo出每個日期的日期在那個特定的一天，在頂部
2. 顯示器所顯示的條目數爲每個用戶的所有數據用戶旁邊

3. 對於每個日期，在數據的上述2個比特的底部，顯示用戶/ s的條目的最高數目

   $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++; 
   
   } 
   

我有麻煩的是，當我嘗試和減少這些子查詢和使用MAX它只會輸出最高計數但不是所有的每個日期，這正是我需要的，包括用戶的輸出/ s的數據這一天的最多條目。

Answer 1

你可以像這樣的東西開始。請注意，我沒有使用PHP的mysql_ API，因爲它已在3年或4年前棄用...

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);