如何使用 MYSQL 和 PHP 将表格数据从垂直变为水平

Question

请帮我解决 2 个表的问题：

1. tbl_data
2. tbl_shipped

tbl_data：

id_data	nama	category
110	Singkong	Asin
111	Ubi	Pedas
112	Pisang	Manis
113	Bawang	Asin
114	Tempe	Pedas Manis
115	Singkong	Pedas
116	Singkong	Manis
117	Singkong	Pedas Manis
118	Ubi	Pedas Manis
119	Ubi	Asin
120	Pisang	Pedas

tbl_shipped：

id	id_data	date	amount
1	110	2021-05-19	40
2	111	2021-05-21	31
3	112	2021-05-29	80
4	112	2021-06-19	50
5	112	2021-06-22	5
6	112	2021-07-30	23
7	113	2021-08-05	70
8	113	2021-08-13	28
9	114	2021-09-04	8
10	115	2021-11-10	67
11	115	2021-12-20	1
12	117	2021-05-19	12
13	117	2021-06-22	50
14	117	2021-08-05	21
15	120	2021-09-04	71
16	120	2021-11-10	53

预期结果：

nama	category	2021-05-19	2021-05-21	2021-05-29	2021-06-19	2021-06-22	2021-07-30	2021-08-05	2021-08-13	2021-09-04	2021-11-10	2021-12-20	Total
Singkong	Asin	40											40
Ubi	Pedas		31										31
Pisang	Manis			80	50	5	23						158
Bawang	Asin							70	28				98
Tempe	Pedas Manis									8			8
Singkong	Pedas										67	1	68
Singkong	Pedas Manis	12				50		21					83
Pisang	Pedas									71	53		60
	All Total :	52	31	80	50	55	23	91	28	15	120	1	546

如果预期结果是上图表格中的枢轴，怎么办？

这是我尝试过但不是最佳的代码（来自请求）：

<table>
<thead>
    <?php include "conn.php";
        $result = $connect->query("SELECT * FROM tbl_shipped LEFT JOIN tbl_data ON tbl_data.id_data = tbl_shipped.id_data ORDER BY tbl_shipped.date ASC"); // query-in database
        $data = array();
        $dates = array();
        while ($row = $result->fetch_array()) { // fetching result
            if(!isset($data[$row['id_data']][$row['date']])) {
                $data[$row['id_data']][$row['date']] = array();
            }     
        
                if(!in_array($row['date'], $dates)) {
                    $dates[] = $row['date'];
                }
            
            $data[$row['id_data']][$row['date']] = $row['amount'];
        }
    ?>
    <tr>
        <th></th>
        <?php foreach($dates AS $date) : ?>
        <th><?=$date?></th>
        <?php endforeach; ?>
    </tr>                                            
</thead>
<tbody>
    <?php foreach($data AS $id => $entries) : ?> 
    <tr>     
        <td><?=$id?></th>                                                              
        <?php foreach($dates AS $date) : ?>
        <td><?=(isset($data[$id][$date])?$data[$id][$date]:'')?></td>
        <?php endforeach; ?>
    </tr>
    <?php endforeach; ?>
</tbody>
<tfoot>

</tfoot>
</table>

根据上面的源代码，结果如下：

	2021-05-19	2021-05-21	2021-05-29	2021-06-19	2021-06-22	2021-07-30	2021-08-05	2021-08-13	2021-09-04	2021-11-10	2021-12-20
110	40
111		31
112			80	50	5	23
113							70	28
114									8
115										67	1
117	12				50		21
120									7	53

请帮助我。我应该如何编写添加代码，我希望得到预期的结果。谢谢。

Answer 1

我至少可以想到一种使用prepared statement 的方法。考虑这个选项：

SET @sql := NULL;
SET @columns := NULL;

SET @columns := (SELECT GROUP_CONCAT(
                                 CONCAT("SUM(CASE WHEN `date`='",dates,"' 
                                                  THEN amount 
                                                   ELSE 0 END) AS '",dates,"'") SEPARATOR ',') 
                  FROM (SELECT DISTINCT `date` AS dates FROM tbl_shipped) dt);

SET @sql := CONCAT("SELECT td.id_data,
                           td.nama,
                            td.category,",@columns,"
                      FROM tbl_shipped AS ts 
                      JOIN tbl_data AS td ON ts.id_data=td.id_data
                  GROUP BY td.id_data, td.nama, td.category;");

PREPARE stmt FROM @sql;
EXECUTE stmt;
DEALLOCATE PREPARE stmt;

我使用@sql 作为最终查询结构的变量，使用@columns 作为列结构的变量。它可以只是一个变量，但我认为这更易于阅读和编辑。

Demo fiddle