如何从表中选择具有相同查询但条件不同的数据库中的数据？

问题描述：

I have two tables:

student
attendance

I want to select the attendance of students by conditions of year and exam_type (midterm, final) and show in one table.

$student_attendance1 = mysqli_query($con, "SELECT * 
FROM student_attendance 
INNER JOIN student 
ON student.student_id = student_attendance.student_id 
WHERE attendance_year=$attendance_year 
AND exam_type=1");

$row_studend_attendance1 = mysqli_fetch_assoc($student_attendance1);

$studend_attendance2 = mysqli_query($con, "SELECT * 
FROM student_attendance 
INNER JOIN student 
ON student.student_id = student_attendance.student_id 
WHERE attendance_year=$attendance_year 
AND exam_type=1");

$row_studend_attendance2 = mysqli_fetch_assoc($student_attendance2);

How can I do this? DATA

我有两个表： p>

学生 li>

出勤率 li> ol>

我想选择 code>学生按年份和考试类型（期中考试，期末考试）的出勤情况并显示在一个表格中。 p>

  $ student_attendance1 = mysqli_query（$ con，“SELECT * 
FROM student_attendance 
INNER JOIN student 
ON student.student_id = student_attendance.student_id 
WHERE  attendance_year = $ attendance_year 
AND exam_type = 1“）; 
 
 $ row_studend_attendance1 = mysqli_fetch_assoc（$ student_attendance1）; 
 
 $ studend_attendance2 = mysqli_query（$ con，”SELECT * 
FROM student_attendance 
INNER JOIN student 
ON student  .student_id = student_attendance.student_id 
WHERE attendance_year = $ attendance_year 
AND exam_type = 1“）; 
 
 $ row_studend_attendance2 = mysqli_fetch_assoc（$ student_attendance2）; 
  code>  pre> 
 
  我怎样才能做到这一点？   DATA   p> 
  div>

答

I will suppose you always have one, and only one, row per student for exam_type = 1 or exam_type = 2(?) (By the way, your two queries, as you wrote them, are absolutely identical...)

You should add a clause like "ORDER BY student.student_id ASC" to be sure you retrieve the datas from your two queries in the same order.

Then, all you have to do is display your datas into a table :

<?php
$student_attendance1 = mysqli_query($con, "SELECT * 
FROM student_attendance 
INNER JOIN student 
ON student.student_id = student_attendance.student_id 
WHERE attendance_year=$attendance_year 
AND exam_type=1 ORDER BY student.student_id ASC");

$studend_attendance2 = mysqli_query($con, "SELECT * 
FROM student_attendance 
INNER JOIN student 
ON student.student_id = student_attendance.student_id 
WHERE attendance_year=$attendance_year 
AND exam_type=2 ORDER BY student.student_id ASC");

while($row_studend_attendance1 = mysqli_fetch_assoc($student_attendance1)) {
    $row_studend_attendance2 = mysqli_fetch_assoc($student_attendance2);
?> // Close your PHP tag.

    <table>
      <tr>
        <th>ID</th>
        <th>Name</th>
        <th>F/Name</th>
        <th>Exams</th>
        <th>Year days</th>
        <th>Present</th>
        <th>Absent</th>
        <th>Sickness</th>
        <th>Permission</th>
      </tr>
      <tr>
        <td rowspan="3"><?= $row_studend_attendance1['student_id']  ?></td>
        <td rowspan="3"><?= $row_studend_attendance1['surname']  ?></td>
        <td rowspan="3"><?= $row_studend_attendance1['firstname']  ?></td>
        <td>Midterm</td>
        <td><?= $row_studend_attendance1['year_days']  ?></td>
        <td><?= $row_studend_attendance1['present']  ?></td>
        <td><?= $row_studend_attendance1['absent']  ?></td>
        <td><?= $row_studend_attendance1['sickness']  ?></td>
        <td><?= $row_studend_attendance1['permission']  ?></td>
      </tr>
      <tr>
        <td>Final</td>
        <td><?= $row_studend_attendance2['year_days']  ?></td>
        <td><?= $row_studend_attendance2['present']  ?></td>
        <td><?= $row_studend_attendance2['absent']  ?></td>
        <td><?= $row_studend_attendance2['sickness']  ?></td>
        <td><?= $row_studend_attendance2['permission']  ?></td>
      </tr>
      <tr>
        <td>Sum</td>
        <td><?= $row_studend_attendance1['year_days'] + $row_studend_attendance2['year_days']  ?></td>
        <td><?= $row_studend_attendance1['present'] + $row_studend_attendance2['present']  ?></td>
        <td><?= $row_studend_attendance1['absent'] + $row_studend_attendance2['absent']  ?></td>
        <td><?= $row_studend_attendance1['sickness'] + $row_studend_attendance2['sickness']  ?></td>
        <td><?= $row_studend_attendance1['permission'] + $row_studend_attendance2['permission']  ?></td>
      </tr>
    </table>

<?php } ?>

I hope I answer your question. If not, feel free to ask again.

如何从表中选择具有相同查询但条件不同的数据库中的数据？

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