从MySQL链接表创建PHP二维数组

Question

我希望有人可以帮我解决这个问题。

我有2张桌子：

课程

--resolve <host:port:address>

文件

+------------+--------------+
| Field      | Type         |
+------------+--------------+
| id         | int(11)      |
| name       | varchar(255) |
+------------+--------------+

我想要的代码返回一个二维PHP数组，如：

+------------+--------------+
| Field      | Type         |
+------------+--------------+
| id         | int(11)      |
| course_id  | int(11)      |
| name       | varchar(255) |
+------------+--------------+

我使用php PDO。我有以下代码：

Array
(
    [0] => Array
        (
            [id] => 3
            [name] => Digital image processing
            [files] => Array
                (
                    [0] => Array
                        (
                            [id] => 6
                            [name] => DIP-ch02-93-1.pdf
                        )

                    [1] => Array
                        (
                            [id] => 9
                            [name] => A_brief_introduction_to_matlab.pdf
                        )

                )

        )

    [1] => Array
        (
            [id] => 1
            [name] => Artificial intelligence
            [files] => Array
                (
                    [0] => Array
                        (
                            [id] => 12
                            [name] => DIP-ch02-93-1.pdf
                        )

                    [1] => Array
                        (
                            [id] => 1
                            [name] => AI-ch03-922.pdf
                        )

                    [2] => Array
                        (
                            [id] => 2
                            [name] => AI-ch04-932.pdf
                        )

                )

        )

)

但它将这个数组归还给我：

$conn = new PDO("mysql:host=$servername;dbname=$dbname", $username, $password);
$stmt = $conn->prepare("SELECT c.id, c.name, f.id as fid, f.name as fname FROM courses c left outer join files f on c.id=f.course_id");
$stmt->execute();

// set the resulting array to associative
$result = $stmt->setFetchMode(PDO::FETCH_ASSOC);
$courses = $stmt->fetchAll();

print_r($courses);

我也有以下正确的代码：

Array
(
    [0] => Array
        (
            [id] => 1
            [name] => Artificial intelligence
            [fid] => 1
            [fname] => AI-ch03-922.pdf
        )

    [1] => Array
        (
            [id] => 1
            [name] => Artificial intelligence
            [fid] => 2
            [fname] => AI-ch04-932.pdf
        )

    [4] => Array
        (
            [id] => 3
            [name] => Digital image processing
            [fid] => 6
            [fname] => DIP-ch02-93-1.pdf
        )

    [5] => Array
        (
            [id] => 3
            [name] => Digital image processing
            [fid] => 9
            [fname] => A_brief_introduction_to_matlab.pdf
        )

    [6] => Array
        (
            [id] => 1
            [name] => Artificial intelligence
            [fid] => 12
            [fname] => DIP-ch02-93-1.pdf
        )

)

它返回正确的结果但我想从我的代码中删除for循环。实际上，我想用单个查询从Mysql中检索数据 有谁可以帮助我？

Answer 1

您需要再次格式化数组。 PDO查询以行方式返回结果。

$conn = new PDO("mysql:host=$servername;dbname=$dbname", $username, $password);
$stmt = $conn->prepare("SELECT c.id, c.name, f.id as fid, f.name as fname FROM courses c left outer join files f on c.id=f.course_id");
$stmt->execute();

// set the resulting array to associative
$result = $stmt->setFetchMode(PDO::FETCH_ASSOC);
$courses = $stmt->fetchAll();

$formattedcourses = array();

foreach($courses as $course){
     $formattedcourses[$course['id']]['id'] = $course['id'];
     $formattedcourses[$course['id']]['name'] = $course['name'];
     $formattedcourses[$course['id']]['files'][] = array(
                  'id' =>  $course['fid'],
                  'name' => $course['fname']
     );
}

Answer 2

SELECT * 
FROM files AS f 
JOIN courses AS c ON c.id = f.course_id
WHERE course_id IN (Select id from courses)"

如果您没有课程条件，则最后一行没有意义 只需在子选择中添加WHERE即可。