PHP 在 while 循环中求和一个值，但有条件

Question

我有两个表要加入，1个是用户，1个是考勤。

TABLE : attendance
id   userId   totalHours 
1    1        0745
2    3        0845
3    1        0945

TABLE : user
id   name  departmentId
1    John  2
2    Sean  2
3    Allan 2

不是每个用户都有出勤记录(他们的总小时数) 但是我需要通过 userId WHERE departmentId = XXXX 和 SUM 他们存在的每个 totalHours 来查询，而不忽略没有任何出勤记录的 userId。

到目前为止我做了这个:

$result = mysqli_query($con,"SELECT * FROM user WHERE departmentId = 2");
while($row = mysqli_fetch_array($result))
{
  $id = $row['userId'];
  $result2 = mysqli_query($con,"SELECT * FROM attendance WHERE userId = $id");
  while($row2 = mysqli_fetch_array($result2))
  $totalHours = 0;
  {
     $totalHours = $row2['totalHours'];
     $grandTotal += $totalHours;
     $totalHoursInHHmm = substr_replace($totalHours,":",2,0); 
     $parsed = date_parse($totalHoursInHHmm); 
     $toSeconds = $parsed['hour'] * 3600 + $parsed['minute'] * 60;
     $total += $toSeconds;
     $init = $total;
     $hours = floor($init / 3600);
     $minutes = floor(($init / 60) % 60);    
  }
  echo "$hours:$minutes";
}

结果显示了部门内的所有用户，并对每个 userId 的所有 totalHours 进行了 SUM，但错误的是，没有任何出勤的 userId 仍然显示 SUM 值，继承了之前的总和

感谢任何帮助:)

Answer 1

I need to query by userId WHERE departmentId = XXXX and SUM each of their totalHours that exist, without neglecting the userId without any record in attendance.

要显示给定部门中所有用户的小时数，甚至是 attendance 表中没有行的用户，请使用 LEFT JOIN

使用 (CAST(totalHours AS UNSIGNED) % 100)/60 + FLOOR(CAST(totalHours AS UNSIGNED)/100) 将您的 varchar 小时+分钟转换为单个小时数。

$query = "SELECT u.id, 
SUM((CAST(totalHours AS UNSIGNED) % 100)/60 + FLOOR(CAST(totalHours AS UNSIGNED)/100)) grandTotal
FROM user u
LEFT JOIN attendance a
ON u.id = a.userId
WHERE u.departmentId = 2
GROUP BY u.id";

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

while($row = mysqli_fetch_array($result)) {
    print $row['id'] . ' ' . $row['grandTotal'];
}