Question

我写了这个查询：

SELECT p.category, r.id, r.code, u.hash_name, r.created_at, r.updated_at FROM uploads AS u
JOIN refrigerators AS r ON u.id=r.image
JOIN products_refrigerators AS pr ON pr.refrigerator_id=r.id
JOIN products p ON pr.product_id = p.id

这就是结果：

Array
(
    [0] => stdClass Object
        (
            [category] => lamps
            [id] => 1
            [code] => canopy
            [path] => public/uploads/AAmAnnC2Yoij9QblcybTJefE5vTQrNZsfDCWqOPt.jpeg
            [created_at] => 2018-02-21 11:15:06
            [updated_at] => 2018-02-21 11:15:06
        )

    [1] => stdClass Object
        (
            [category] => panels
            [id] => 1
            [code] => canopy
            [path] => public/uploads/AAmAnnC2Yoij9QblcybTJefE5vTQrNZsfDCWqOPt.jpeg
            [created_at] => 2018-02-21 11:15:06
            [updated_at] => 2018-02-21 11:15:06
        )

    [2] => stdClass Object
        (
            [category] => lamps
            [id] => 2
            [code] => canopy2
            [path] => public/uploads/nqDT4gEXbF0opPmEpa1cPo5iPSulRoesYcgHiOA3.jpeg
            [created_at] => 2018-02-21 11:17:44
            [updated_at] => 2018-02-21 11:17:44
        )

    [3] => stdClass Object
        (
            [category] => lamps
            [id] => 3
            [code] => QWE
            [path] => public/uploads/0TSrz6j3JJH17XgEs2161mOh8aYgDikujdYwrCWR.jpeg
            [created_at] => 2018-02-21 11:18:13
            [updated_at] => 2018-02-21 11:18:13
        )

    [4] => stdClass Object
        (
            [category] => panels
            [id] => 3
            [code] => QWE
            [path] => public/uploads/0TSrz6j3JJH17XgEs2161mOh8aYgDikujdYwrCWR.jpeg
            [created_at] => 2018-02-21 11:18:13
            [updated_at] => 2018-02-21 11:18:13
        )

)

我的问题是：有没有办法得到这样的结果（=将category属性转换为一个数组，如果存在多个相关类别，则对其进行分组）

Array
(
    [0] => stdClass Object
        (
            [category] => Array 
                          (
                              [0] => lamps
                              [1] => panels
                          )
            [id] => 1
            [code] => canopy
            [path] => public/uploads/AAmAnnC2Yoij9QblcybTJefE5vTQrNZsfDCWqOPt.jpeg
            [created_at] => 2018-02-21 11:15:06
            [updated_at] => 2018-02-21 11:15:06
        )

    [1] => stdClass Object
        (
            [category] => Array 
                          (
                              [0] => lamps
                          )
            [id] => 2
            [code] => canopy2
            [path] => public/uploads/nqDT4gEXbF0opPmEpa1cPo5iPSulRoesYcgHiOA3.jpeg
            [created_at] => 2018-02-21 11:17:44
            [updated_at] => 2018-02-21 11:17:44
        )

    [2] => stdClass Object
        (
            [category] => Array 
                          (
                              [0] => lamps
                              [1] => panels
                          )
            [id] => 3
            [code] => QWE
            [path] => public/uploads/0TSrz6j3JJH17XgEs2161mOh8aYgDikujdYwrCWR.jpeg
            [created_at] => 2018-02-21 11:18:13
            [updated_at] => 2018-02-21 11:18:13
        )

)

直接使用MySQL或唯一的方法是操纵结果（在PHP中，在我的情况下）？

可能答案是“你必须使用PHP”，但我不是MySQL专家，这就是问题的原因。

Answer 1

你应该连接你的category列，然后按ID分组，就像Patrick在评论中提到的那样。

SELECT GROUP_CONCAT(p.category), r.id ... GROUP BY r.id

您也可以使用JSON_ARRAY或JSON_OBJECT函数来获得类似的结果，但不要忘记将它们分组

查询在SQL或PHP中的结果操作？

1 个答案: