一个Select语句,它将执行以下操作

Question

我正在学习如何围绕sql和php.我有4个表格结构如下

+-----------+    +------------+    +---------+    +----------+
|  Project  |    | Slide      |    | Shape   |    |  Points  |
+-----------+    +------------+    +---------+    +----------+
|    id     |    |  id        |    | id      |    | id       |
+-----------+    | project_id |    | cont_id |    | shape_id |
                 +------------+    +---------+    | x        |
                                                  | y        |
                                                  +----------+

正如您所看到的,表格通过id一直链接到点,这意味着项目将包含许多包含许多包含多个点的形状的幻灯片.

我有一个SQL查询

SELECT slide.`id`, shape.`id`, points.`x_point`, points.`y_point` 
FROM `project`, `slide`, `shape`, `points` 
WHERE 1 = slide.`project_id` 
   AND slide.`id` = shape.`slide_id` 
   AND shape.`id` = points.`shape_id`

我想要的是将此查询的结果看起来像这样

[0] => stdClass Object
     (
         [id] => 27
         [x] => 177
         [y] => 177
     )

 [1] => stdClass Object
     (
         [id] => 27
         [x] => 178
         [y] => 423
     )

 [2] => stdClass Object
     (
         [id] => 27
         [x] => 178
         [y] => 419
     )

 [3] => stdClass Object
     (
         [id] => 27
         [x] => 178
         [y] => 413
     )

 [4] => stdClass Object
     (
         [id] => 27
         [x] => 181
         [y] => 399
     )

 [5] => stdClass Object
     (
         [id] => 27
         [x] => 195
         [y] => 387
     )

 [6] => stdClass Object
     (
         [id] => 27
         [x] => 210
         [y] => 381
     )

 [7] => stdClass Object
     (
         [id] => 27
         [x] => 231
         [y] => 372
     )

 [8] => stdClass Object
     (
         [id] => 27
         [x] => 255
         [y] => 368
     )

 [9] => stdClass Object
     (
         [id] => 27
         [x] => 283
         [y] => 368
     )
... AND CONTINUED FOR A LONG TIME

我想要的是将这个野蛮的垃圾堆转换成更像这样的东西

[9] => stdClass Object
         (
             [id] => ID OF LIKE SHAPES
             [x] => Array(ALL THE X POINTS)
             [y] => ARRAY(ALL THE Y Points)
         )

我不能为我的生活弄清楚如何将其转换为这样的数组.

如果用我设计的查询无法完成,那么有更好的查询.也许一个抓住积分,然后把它放到一个点数组......我想我只是有一个想法......

---

新信息,

所以我在这个问题上添加了答案,我不知道这是否是标准方式.如果我不是一个好的解决方案,为了帮助解决其他问题,我也会在这里添加我的思考过程.

有关详细信息,请查看我的答案.

ORM如何与我的算法进行比较？

Answer 1

使用像Doctrine这样的 ORM ，你可以简单地对其进行建模

/**
 * @Entity
 */
class Project
{
    /**
     * @Id @GeneratedValue
     * @Column(type="integer")
     */
    private $id;

    /**
     * @OneToMany(targetEntity="Slide", mappedBy="project")
     */
    private $slides;

    public function __construct()
    {
        $this->slides = new \Doctrine\Common\Collections\ArrayCollection;
    }
}

/**
 * @Entity
 */
class Slide
{
    /**
     * @Id @GeneratedValue
     * @Column(type="integer")
     */
    private $id;

    /**
     * @ManyToOne(targetEntity="Project", inversedBy="slides")
     * @JoinColumn(name="project_id", referencedColumnName="id")
     */
    private $project;

    /**
     * @OneToMany(targetEntity="Shape", mappedBy="slide")
     */
    private $shapes;
}

等等...

请参阅http://www.doctrine-project.org/docs/orm/2.0/en/reference/association-mapping.html#one-to-many-bidirection

当然，这涉及相当多的设置和处理开销，但随着域模型变得更加复杂，您会欣赏 ORM。