Warning: file_get_contents(/data/phpspider/zhask/data//catemap/1/php/269.json): failed to open stream: No such file or directory in /data/phpspider/zhask/libs/function.php on line 167

Warning: Invalid argument supplied for foreach() in /data/phpspider/zhask/libs/tag.function.php on line 1116

Notice: Undefined index: in /data/phpspider/zhask/libs/function.php on line 180

Warning: array_chunk() expects parameter 1 to be array, null given in /data/phpspider/zhask/libs/function.php on line 181

Warning: file_get_contents(/data/phpspider/zhask/data//catemap/4/r/79.json): failed to open stream: No such file or directory in /data/phpspider/zhask/libs/function.php on line 167

Warning: Invalid argument supplied for foreach() in /data/phpspider/zhask/libs/tag.function.php on line 1116

Notice: Undefined index: in /data/phpspider/zhask/libs/function.php on line 180

Warning: array_chunk() expects parameter 1 to be array, null given in /data/phpspider/zhask/libs/function.php on line 181
Php 存储Post和用户id,参考Laravel 6中的用户表_Php_Eloquent_Model_Relationship_Laravel 6 - Fatal编程技术网

Php 存储Post和用户id,参考Laravel 6中的用户表

Php 存储Post和用户id,参考Laravel 6中的用户表,php,eloquent,model,relationship,laravel-6,Php,Eloquent,Model,Relationship,Laravel 6,我的帖子和用户模型的关系如下: user\u id是posts表中引用users表的id的外键 后模型 用户模型 以及具有存储方法的后置控制器: public function store(Request $request){ $this->validate($request, [ 'title' => 'required|max:255', 'desc' => 'required|max:255' ]); $image = null; $files = $r

我的帖子和用户模型的关系如下:
user\u id
posts
表中引用
users
表的
id
的外键 后模型

用户模型

以及具有存储方法的
后置控制器

public function store(Request $request){
 $this->validate($request, [
  'title' => 'required|max:255',
  'desc' => 'required|max:255'
 ]);

 $image = null;
 $files = $request->thumbnail;
 $thumb = time().'.'.$files->getClientOriginalExtension();
 $request->thumbnail->move(public_path('public'),$thumb);
 $image = $thumb;
 $post = Posts::create([
   'title' => $request->title,
   'thumbnail' => $image,
   'user_id' => __________
   'desc' => $request->desc
 ]);
 return $post->toJson();                     }
考虑到这一点,我想建立一种关系:一个用户可以有很多帖子。 那么,如何借助
用户
posts
表之间的关系,获取
用户
id
,以插入posts表中
编辑:在上面的控制器中,如何获取用户id值???

非常简单

您可以通过url获取id:

localhost:8000/users/{id}
在您的路线中[web.php]

Route::get('/users/{id}', 'UserController@show');
您的控制器:

<?php

namespace App\Http\Controllers;


class UserController extends Controller
{
    public function show($id)
    {
        return User::with('posts')->where('id',$id)->get()
    }
不要忘记名称空间

use Illuminate\Support\Facades\Auth;

return User::with('posts')->where('id',$id)->get()对不起,我想我没有提到这一点,我需要获取用户id以存储在posts表中。我编辑了问题'user\u id'=>Auth::user()->id;
<?php

namespace App\Http\Controllers;


class UserController extends Controller
{
    public function show($id)
    {
        return User::with('posts')->where('id',$id)->get()
    }
// Get the currently authenticated user...
$user = Auth::user();

// Get the currently authenticated user's ID...
$id = Auth::id();
use Illuminate\Support\Facades\Auth;