如何在Laravel雄辩中建立三个表之间的关系？
问题描述：
我试图从表中检索数据，我没有从我目前使用的模型直接关系。如何在Laravel雄辩中建立三个表之间的关系？
我的数据结构：
表：帖子
- ID - 整数
- 标题 - 字符串
表：post_stacks
- ID - 整数
- POST_ID - 整数
- stack_id - 整数
表：栈
- ID - 整数
- 体 - 字符串
- 网址 - 串
我的口才模型是从post.php中（职位表），我试图让关联到我的帖子的所有堆栈（从栈表）。我只想在Post.php而不是我的数据透视表（post_stacks）上声明我的关系。
我试过hasManyThrough，但我的表结构不符合Laravel要求的需求，因为我需要在我的Stacks表上使用外键。
这是我目前的执行：
post.php中
class Post extends Model
{
protected $dates = [
'created_at',
'updated_at'
];
public function user()
{
return $this->belongsTo(\App\User::class, 'user_id', 'id');
}
public function post_stacks()
{
return $this->hasMany(\App\PostStack::class);
}
public function post_os()
{
return $this->hasMany(\App\PostOS::class, 'post_id', 'id');
}
public function post_tags()
{
return $this->hasMany(\App\PostTag::class , 'post_id', 'id');
}
public function getCreatedAtAttribute($value)
{
return Carbon::parse($value)->toFormattedDateString();
}
}PostController.php
class PostController extends Controller
{
public function index()
{
$posts = Post::all();
foreach($posts as $post){
$post->user;
$post->created_at;
$posts_os = $post->post_os;
$post_stacks = $post->post_stacks;
$post_tags = $post->post_tags;
foreach($posts_os as $post_os){
$os = OS::where('id', $post_os->os_id)->first();
$post_os['body'] = $os['body'];
}
foreach($post_stacks as $post_stack){
$stack = Stack::where('id', $post_stack->stack_id)->first();
$post_stack['url'] = $stack['url'];
$post_stack['body'] = $stack['body'];
}
foreach($post_tags as $post_tag){
$tag = Tag::where('id', $post_tag->tag_id)->first();
$post_tag['body'] = $tag['body'];
}
}
return response()->json($posts);
}
}我的JSON数据响应
[
{
"id":1,
"title":"Laravel + XAMPP",
"user_id":1,
"description":"I'll take you through the entire process of setting up a development environment for Laravel using XAMPP.",
"created_at":"Jun 12, 2017",
"updated_at":"2017-06-12 08:55:02",
"user":{
"id":1,
"name":"EpIEhg7ciO",
"email":"[email protected]",
"created_at":"2017-06-12 08:55:02",
"updated_at":"2017-06-12 08:55:02"
},
"post_os":[
{
"id":1,
"post_id":1,
"os_id":1,
"created_at":"2017-06-12 08:55:02",
"updated_at":"2017-06-12 08:55:02",
"body":"Windows"
}
],
"post_stacks":[
{
"id":1,
"post_id":1,
"stack_id":1,
"created_at":"2017-06-12 08:55:02",
"updated_at":"2017-06-12 08:55:02",
"url":"laravel.svg",
"body":"Laravel"
},
{
"id":2,
"post_id":1,
"stack_id":2,
"created_at":"2017-06-12 08:55:02",
"updated_at":"2017-06-12 08:55:02",
"url":"xampp.svg",
"body":"XAMPP"
}
],
"post_tags":[
{
"id":1,
"post_id":1,
"tag_id":1,
"created_at":"2017-06-12 08:55:02",
"updated_at":"2017-06-12 08:55:02",
"body":"laravel"
},
{
"id":2,
"post_id":1,
"tag_id":2,
"created_at":"2017-06-12 08:55:02",
"updated_at":"2017-06-12 08:55:02",
"body":"xampp"
}
]
},
{
"id":2,
"title":"Laravel + Vagrant",
"user_id":1,
"description":"I'll take you through the entire process of setting up a development environment for Laravel using Vagrant.",
"created_at":"Jun 12, 2017",
"updated_at":"2017-06-12 08:55:02",
"user":{
"id":1,
"name":"EpIEhg7ciO",
"email":"[email protected]",
"created_at":"2017-06-12 08:55:02",
"updated_at":"2017-06-12 08:55:02"
},
"post_os":[
{
"id":2,
"post_id":2,
"os_id":1,
"created_at":"2017-06-12 08:55:02",
"updated_at":"2017-06-12 08:55:02",
"body":"Windows"
},
{
"id":3,
"post_id":2,
"os_id":2,
"created_at":"2017-06-12 08:55:02",
"updated_at":"2017-06-12 08:55:02",
"body":"Mac OS X"
},
{
"id":4,
"post_id":2,
"os_id":3,
"created_at":"2017-06-12 08:55:02",
"updated_at":"2017-06-12 08:55:02",
"body":"Linux"
}
],
"post_stacks":[
{
"id":3,
"post_id":2,
"stack_id":1,
"created_at":"2017-06-12 08:55:02",
"updated_at":"2017-06-12 08:55:02",
"url":"laravel.svg",
"body":"Laravel"
},
{
"id":4,
"post_id":2,
"stack_id":3,
"created_at":"2017-06-12 08:55:02",
"updated_at":"2017-06-12 08:55:02",
"url":"vagrant.png",
"body":"Vagrant"
}
],
"post_tags":[
{
"id":3,
"post_id":2,
"tag_id":1,
"created_at":"2017-06-12 08:55:02",
"updated_at":"2017-06-12 08:55:02",
"body":"laravel"
},
{
"id":4,
"post_id":2,
"tag_id":3,
"created_at":"2017-06-12 08:55:02",
"updated_at":"2017-06-12 08:55:02",
"body":"vagrant"
}
]
}
]我的JSON数据是正是我想要的。我只是觉得我的PostController实现效率低下，并且运行了太多的查询。我运行的查询太多，并且有嵌套循环。有没有一种干净的方式可以使用Laravel的方法/关系之一建立关系？
谢谢！
答
你可以在栈模型上声明关系吗？您只排除在数据透视表上添加关系。如果是这样的：
class Post extends Model
{
public function stacks()
{
return $this->hasMany(\App\Stack::class);
}
}
class Stack extends Model
{
public function posts()
{
return $this->belongsToMany(\App\Post::class);
}
}
通过叠后模型中定义的字段，这应该是所有你需要的雄辩，使关系的工作。此外，您可以访问pivot属性，让您：
$post->pivot->somePropertyOnStack
编辑从文档
摘录给你如何雄辩确定关系的总体思路：
请记住，Eloquent将自动确定注释模型上正确的外键列。按照惯例，Eloquent将采用拥有模型的“蛇情况”名称并以_id作为后缀。因此，对于这个例子，Eloquent会假设Comment模型中的外键是post_id。
并与连接表的关系：
确定了恋爱关系的连接表的表名，机锋将加入按字母顺序排列的两个相关的型号名称。
的文档的本节将介绍每个关系类型和雄辩如何发生他们：
https://laravel.com/docs/5.4/eloquent-relationships#defining-relationships
我不太明白，你是如何与这两种模式？有没有外键建立关系？ – 75Kane
通过表名和你定义的字段。当你将它们添加到模型类时，雄辩会假设一些关于表之间关系的约定。 – btl
请记住，这只适用于您的中间表名称字段名称。他们完全遵循雄辩的期望，所以你不需要“指导”如何建立关系的雄辩。 – btl