原则：如何插入外键值 [英] Doctrine: How to insert foreign key value

问题描述

我的数据库中有两张表：滑块和图像。一个滑块可以有很多图像，所以表的结构是：

I have two tables in my database: sliders and images. One slider can have many images, so the structure of tables is:

 ---------      --------
| SLIDERS |    | IMAGES |
 ---------      --------
|   id    |    |   id   |
 ---------      --------
|  title  |    | title  |
 ---------      --------
               |  sid   |
                --------
SID is a foreign key associated to id in "SLIDERS" table.

在实体中，我将双向关系 OneToMany 和 ManyToOne ，所以获取的Slider结果将包含属于他的 Images Objects ，而图像将包含 Slider Object 属于他们。

In entities I put bidirectional relationships OneToMany and ManyToOne, so fetched Slider result would contain Images Objects that belong to him, and Images would contain Slider Object that belongs to them.

滑块实体：

class Sliders
{

...

/**
 * @ORM\OneToMany(targetEntity="Images", mappedBy="slider")
 */
protected $images;

图片实体：

class Images
{

...

/**
 * @ORM\ManyToOne(targetEntity="Sliders", inversedBy="images")
 * @ORM\JoinColumn(name="sid", referencedColumnName="id")
 */
protected $slider;

提取真的很舒服。但是现在我不明白如何将 INSERT sid into Images 表，因为我的实体中没有这个字段，除了 $ slider 返回Object。有可能使用这个 $ slider ？

It is really comfortable for fetching. But now I can't understand how can I INSERT sid into Images table, since I don't have this field in my entity except $slider that returns Object. Is it possible to do it using this $slider?

推荐答案

功能你应该已经在你的Slider实体（或类似的）。

Following function you should have already in your Slider entity (or similar).

public function addImage(Image $image)
{
    $image->setSlider($this); // This is the line you're probably looking for
    $this->images[] = $image;

    return $this;
}

它的作用是如果你坚持写实体的ID （sid）进入您的图像。

What it does is if you persist the entity it writes the ID of the Slider (sid) into your Image.

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