使用PHP从MySQL数据库中获取数据

Question

我有3张表：
1）出勤
2）登记
3）时间表

1。出勤
sl_no是注册sl_no的外键引用，AttID是主键

2。 registeration
sl_no是主键

第3。时间表
id是主键

我正在针对schedule table和registration table查询我的数据库，以获取日程安排开始日期，结束日期和学生详细信息where university is same，如下所示：

<?php   
        $schedule_query = "SELECT a.scheduleStartDate, a.scheduleEndDate, b.sl_no, b.student_name 
        FROM schedule a 
        LEFT JOIN registration b 
        ON a.university = b.university 
        WHERE a.scheduleName = 'XYZ Schedule 01'";

        $result = mysqli_query($link, $schedule_query);
        $row = mysqli_fetch_array($result);
        $start = strtotime($row['scheduleStartDate']);
        $end = strtotime($row['scheduleEndDate']);
        $date = $start;
        $attendance_array = array();
    ?>

我使用以下查询以表格格式显示报告：

    <table width="100%" class="tbl" style="font-size: 12px">
<tr>
<th>Sl No</th>  
<th>Name</th>
<?php
$ttl = '';
while($date <= $end)
{
    $ttl++;
    $student_array = array();
    $stu_name_td = '';
    mysqli_data_seek($result,0);
    //Result came from schedule and registration
    while($innerrow = mysqli_fetch_array($result)){      
        $atd_query = "SELECT * FROM attendance WHERE AttDate = '".date('Y-m-d', $date)."' AND sl_no = '".$innerrow['sl_no']."'";
        //results from attendance
        $present_stu_res = mysqli_query($link, $atd_query);
        $attendance_array[$innerrow['student_name']][date('Y-m-d', $date)] =  mysqli_num_rows($present_stu_res) > 0 ? 1 : "A";
    }
    echo "<th>".date('j/m/Y', $date)."</th>";
    $date = strtotime("+1 day", $date);
 }
?>
<th>Num of Days</th>
<th>Attended</th>
<th>Percentage(%)</th>
</tr>
<?php
$count ='';
    foreach($attendance_array as $stu_name=>$innerarray){
$count ++;
?>
    <tr>
    <td><?php echo $count;?></td>
    <td><?php echo $stu_name; ?></td>
    <?php 
        foreach($innerarray as $dateval=>$present_val)
            echo '<td>'.$present_val.'</td>';
           echo '<td>'.$ttl.'</td>';//Total No of Days
            $attended = array_sum($innerarray);
            echo '<td>'.$attended.'</td>';//No of Days Attended
            $percent_cal = $attended / $ttl; 
            $percent = number_format( $percent_cal * 100, 2 ) . '%'; 
            echo '<td>'.$percent.'</td>';//Percentage
    ?>
    </tr>
    <?php } ?>
<?php }?>
</tr>
</table>

从上面的代码我得到如下输出：

Student Name 17/06/2015   18/06/2015  19/06/2015 20/06/2015 21/06/2015 Num_of_days Attended Percentage(%) 
Student1       1             A          1           A         1             6           3       60%
Student2       1             A          1           A         1             6           3       60%
Student3       1             A          1           A         1             6           3       60%

我的问题是：
我只考虑日期为17日和21日的日期XYZ Schedule 01，我不想在报告中显示，并相应地计算百分比，我该如何实现？任何帮助都可能非常有用。

Answer 1

取出几行代码。你应该拿出的第一行是

<TH>percentage(%)</TH>

您要删除的下两行是

$percent = number_format( $percent_cal * 100, 2 ) . '%'; 
echo '<td>'.$percent.'</td>';//Percentage