Question

我有一个数据库设置，它可以保存已经收集了一定金额的销售人员的数据，并在那里也有佣金。

 Example:
 Date        Name    Amount Sold   Commission Earned
 2013-2-3   John A     100.00          20.00
 2013-2-3   Amy W      200.00          40.00
 2013-2-3   John A     230.00          46.00
 2013-2-2   John A     300.00          60.00
 2013-2-2   Bill B     400.00          80.00
 Etc....

我试图通过整周赢得的佣金来展示最畅销的销售人员。我需要以某种方式查询数据库以查找本周的所有佣金，然后将它们分组并汇总所获得的总数并显示本周的最高收入者。

这是我到目前为止所做的工作，但它显示了上周所有收入。我只是不知道如何按照他们的名字对它们进行分组，然后将佣金加在一起，然后从最赚钱的销售人员那里展示。

   require_once('connectvars.php');
   $dbc = mysqli_connect(CDB_HOST, CDB_USER, CDB_PASSWORD, CDB_NAME);
   // GET COLLECTOR INFO
   $query = "SELECT * FROM commissioninfo WHERE thedate >= curdate() - INTERVAL DAYOFWEEK(curdate())+6 DAY 
   AND thedate < curdate() - INTERVAL DAYOFWEEK(curdate())-1 DAY ORDER BY amountoffeeearned DESC"; 
   $result = mysqli_query($dbc, $query);

     while ($row = mysqli_fetch_array($result)) {
         echo '<tr>';
         echo '<td>' . $row['salesperson'] . '</td>';
         echo ' <td>' . $row['amountoffeeearned'] . '</td>';
         echo '</tr>';                              
      }
      mysqli_close($dbc);

Answer 1

只需在Name中添加Group by，然后Sum Amount Sold，您将获得Amount Sold每user的总和。这样的事情。

SELECT   NAME, 
         Sum(`Amount Sold`)     sum_amount , 
         sum(`commission earned`) sum_comm 
FROM     commissioninfo 
WHERE    thedate >= curdate() - interval dayofweek(curdate())+6 day 
AND      thedate < curdate()  - interval dayofweek(curdate())-1 day 
GROUP BY NAME 
ORDER BY sum_amount DESC

更新：

SELECT salesperson, 
       Sum(amountcollected) 
FROM   commissioninfo 
WHERE  thedate >= Curdate() - INTERVAL Dayofweek(Curdate())+6 day 
       AND thedate < Curdate() - INTERVAL Dayofweek(Curdate())-1 day 
GROUP  BY salesperson

Answer 2

按顺序在DESC中添加分组依据和SUM（[佣金获得]）中的名称。

或

试试这个：

SELECT Name, SUM([Amount Sold]) as WeekAmountSold, SUM([Commission Earned]) as WeekCommissionEarned FROM commissioninfo WHERE Date >= DATE_ADD(curdate(), INTERVAL(1-DAYOFWEEK(curdate())-7) DAY) AND Date < DATE_ADD(curdate(), INTERVAL(1-DAYOFWEEK(curdate())) DAY) GROUP BY Name ORDER BY SUM([Commission Earned]) DESC

PHP查询获取日期之间每个用户的金额的和

2 个答案: