如何在同一天比较多个时间表和出席

Question

我为这个微不足道的问题表示歉意，花了很多小时使用谷歌搜索这个问题的答案。请帮助.. 我有一个学生的日程安排表和学生的表出勤表考勤。我想要的是比较学生的时间表和出席人数。

表时间表

ID | stud_id | day | timein | timeout | schedule_hour | 
1 | 8888 | Monday| 08:00AM | 09:00AM | 1   | 
2 | 8888 | Monday| 10:00AM | 12:00PM | 2   | 
3 | 8888 | Monday| 01:00PM | 03:00PM | 2   | 
4 | 8888 |Tuesday| 07:00AM | 09:00AM | 2   | 
5 | 8888 |Tuesday| 03:00PM | 04:00PM | 1   | 
6 | 8888 |Tuesday| 06:00PM | 08:00PM | 2   | 

表出席

ID | stud_id | day | timein | timeout | hour | 
1 | 8888 | Monday| 07:55AM | 09:10AM |1.15 | 
2 | 8888 | Monday| 09:51AM | 12:03PM |2.12 | 
3 | 8888 | Monday| 12:50PM | 03:05PM |2.15 | 
4 | 8888 |Tuesday| 06:53AM | 09:15AM |2.22 | 
5 | 8888 |Tuesday| 02:50PM | 04:06PM |1.16 | 
6 | 8888 |Tuesday| 05:52PM | 08:09PM |2.17 | 
7 | 8888 |Tuesday| 09:52PM | 10:09PM |1.17 | 

PHP代码如下： -

<?php 
$query=$conn->query("SELECT * 
FROM student a,attendance d,schedule e 
WHERE a.student_id = d.student_id 
AND d.student_id = e.student_id 
AND a.student_id = $get_id")or die(mysql_error()); 
$totalhour = 0; 
$totalitems = 0; 

function hourandmin($time1){ 
$hour = 0; 
$minute = 0; 
while($time1 > 1.0){ 
$hour = $hour + 1; 
$time1 = $time1 - 1; 
} 

if($time1 < 1.0){ 
$minute = $time1 * 60; 
} 

$time1 = round($hour).".".round($minute).""; 
return $time1; 
} 

while($row = $query->fetch()){ 

$date = strtotime($row['login']); 
$date2 = strtotime($row['logout']); 
$hour = ($date2 - $date)/(60*60); 
if ($hour < 0){ 
$hour = 0; 
} 
?> 
<tr> 
<td><?php echo $row['student_id']; ?></td> 
<td><?php echo $row['day']; ?></td> 
<td><?php echo $row['timein']; ?></td> 
<td><?php echo $row['timeout']; ?></td> 
<td><?php echo hourandmin((($hour))); ?></td> 
<td><?php echo $schedule_hour; ?></td> 





           </tr> 
            <?php } ?> 

我不知道如何输出的时间表小时 因此，例如在他出席的第7排他计时，但因为他没有时间表在此时的输出是加班

ID | stud_id | day | timein | timeout | hour |schedule_hour 
1 | 8888 | Monday| 07:55AM | 09:10AM | 1.15| 1 
2 | 8888 | Monday| 09:51AM | 12:03PM | 2.12| 2 
3 | 8888 | Monday| 12:50PM | 03:05PM | 2.15| 2 
4 | 8888 |Tuesday| 06:53AM | 09:15AM | 2.22| 2 
5 | 8888 |Tuesday| 02:50PM | 04:06PM | 1.16| 1 
6 | 8888 |Tuesday| 05:52PM | 08:09PM | 2.17| 2 
7 | 8888 |Tuesday| 09:52PM | 10:09PM | 1.17| overtime 

Answer 1

希望这有助于

SELECT tatt.stud_id, tatt.day, tatt.timein, tatt.timeout, tatt.hour,tsched.schedule_hour FROM table_attendance tatt INNER JOIN table_schedule tsched ON tsched.stud_id = tatt.stud_id AND tsched.day = tatt.day AND tatt.timein <= tsched.timein AND tatt.timeout > tsched.timeout; 

Answer 2

<?php 
$query=$conn-> 
query(" 
select a.student_id, d.timein, d.timeout, e.timein, e.timeout 
from student a 
inner join attendance d on a.student_id = d.student_id 
inner join schedule e on a.student_id = e.student_id 
where a.student_id = $get_id 
")or die(mysql_error()); 
$totalhour = 0; 
$totalitems = 0; 

function hourandmin($time1){ 
$hour = 0; 
$minute = 0; 
while($time1 > 1.0){ 
$hour = $hour + 1; 
$time1 = $time1 - 1; 
} 

if($time1 < 1.0){ 
$minute = $time1 * 60; 
} 

$time1 = round($hour).".".round($minute).""; 
return $time1; 
} 

while($row = $query->fetch()){ 

$date = strtotime($row['d.login']); 
$date2 = strtotime($row['d.logout']); 

$hour = ($date2 - $date)/(60*60); 
if ($hour < 0){ 
$hour = 0; 
} 

function schedule($date, $date2){ 
$date3 = strtotime($row['e.timein']); 
$date4 = strtotime($row['e.timeout']); 
if $date1 < $date3 && $date2 > $date4 { 
$schedule_hour = $date4 - $date3; 
else { 
$schedule_hour = round($hour); 
} 

} 

?> 
<tr> 
<td><?php echo $row['student_id']; ?></td> 
<td><?php echo $row['day']; ?></td> 
<td><?php echo $row['timein']; ?></td> 
<td><?php echo $row['timeout']; ?></td> 
<td><?php echo hourandmin((($hour))); ?></td> 
<td><?php echo $schedule_hour; ?></td>