我正在尝试从与我目前使用的模型没有直接关系的表中检索数据。
我的数据结构：
表：帖子
表： post_stacks
表：堆叠
我雄辩的模型来自Post.php（posts表），我正在尝试获取与我的帖子相关的所有堆栈（来自堆栈表）。我想仅在Post.php而不是我的数据透视表（post_stacks）上声明我的关系。
我已经尝试hasManyThrough但是我的表的结构不符合Laravel所要求的要求，因为我需要在Stacks表上使用外键。
这是我目前的实施：
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数据回复
[
{
"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":"AiyZXrubVG@gmail.com",
"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":"AiyZXrubVG@gmail.com",
"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数据正是我想要的。我只是认为我的PostController实现效率低下并且运行了太多查询。我运行太多查询并具有嵌套循环。我是否可以使用Laravel的方法/关系之一建立关系？
谢谢！
答案 0 :(得分：2)
你能在Stack模型上声明关系吗？您只排除了在数据透视表上添加关系。如果是这样的话：
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PostStack模型中定义的字段，这应该是Eloquent使关系工作所需的全部内容。此外，您可以访问pivot属性，允许您：
$post->pivot->somePropertyOnStack
修改
摘录文档，让您大致了解雄辩如何决定关系：
请记住，Eloquent会自动确定Comment模型上的正确外键列。按照惯例，Eloquent将采取＆＃34;蛇案例＆＃34;拥有模型的名称，并以_id为后缀。因此，对于此示例，Eloquent将假定Comment模型上的外键是post_id。
与加入表格的关系：
确定关系表的连接表，Eloquent将按字母顺序加入两个相关的模型名称。
这部分文档将解释每种关系类型以及Eloquent如何发生它们：
https://laravel.com/docs/5.4/eloquent-relationships#defining-relationships