如何使用 Laravel 中的帖子查询用户
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How to query user with posts in Laravel
提问by DNM
I'm new to Laravel. I want to display all post including the user who posted the post. I have a simple query which displays all post, and displays selected post using post id.
我是 Laravel 的新手。我想显示所有帖子，包括发布帖子的用户。我有一个简单的查询，它显示所有帖子，并使用帖子 ID 显示选定的帖子。
public function view_post($id)
{
$post = User::find($id);
dd($post);
}
This is working perfectly fine but I wanted to display all post with the user who posted the post. How can I do this in proper way?
这工作得很好，但我想与发布帖子的用户一起显示所有帖子。我怎样才能以正确的方式做到这一点？
回答by Tim Lewis
Right now, it is possible to get the Postsby first getting the Userthen using that ID, getting the Posts. Note: this is not the recommended way of using Laravel with relationships:
现在，可以Posts通过先获取 ，User然后使用该 ID，获取Posts. 注意：这不是将 Laravel 用于关系的推荐方式：
$user = User::find($id);
$posts = Post::where("user_id", "=", $user->id)->get();
While this would work (assuming you had a model for Postwith a user_idkey), but the correct way is to define the relationship in the models. For this one, it would be hasMany()and belongsTo()on Userand Postrespectfully. Here is your User.phpmodel class:
虽然这会起作用（假设您有一个Post带user_id键的模型），但正确的方法是定义模型中的关系。对于这一个，这将是hasMany()与belongsTo()上User和Post尊敬。这是您的User.php模型类：
class User extends Eloquent {
protected $table = "users";
public function posts(){
return $this->hasMany("Post");
}
}
And make sure to define the inverse in your Post.phpmodel class:
并确保在您的Post.php模型类中定义逆：
class Post extends Eloquent {
protected $table = "posts";
public function user(){
return $this->belongsTo("User");
}
}
Lastly, you can query this relationship in your controller using the following:
最后，您可以使用以下命令在控制器中查询此关系：
$user = User::find($id);
$posts = $user->posts()->get();
This will also enforce integrity by forcing any updates/saves to use the right user_idkey. There's some pretty extensive documentation on how to use the Eloquentstyle of database querying, check it out here:
这也将通过强制任何更新/保存使用正确的user_id密钥来强制执行完整性。有一些关于如何使用Eloquent数据库查询样式的非常广泛的文档，请在此处查看：
Hope that helps!
希望有帮助！
Edit
编辑
To pass this to a view, add the following to your controller:
要将其传递给视图，请将以下内容添加到您的控制器中：
public function view_post($id){
$user = User::find($id);
$posts = $user->posts()->get();
return View::make("view")->with(array("user" => $user, "posts" => $posts));
}
You will need a file in your app/viewsdirectory named (in this case) view.blade.phpwhich "echos" all the information about the user and the posts:
您将需要在您的app/views目录中创建一个名为（在本例中）的文件，该文件view.blade.php“回显”有关用户和帖子的所有信息：
<h2>User: {{ $user->id }} - {{ $user->email }}</h2>
<br/>
<h2>Posts:</h2>
@foreach($posts AS $post)
<p> {{ $post->id }}: {{ $post->content }}</p>
@endforeach
Etc etc. Note, I have no idea what columns your usersand poststable has, but using $user->columnor $post->columnwill echo the contents of that named column.
等等等等。注意，我不知道你的users和posts表有哪些列，但使用$user->column或$post->column将回显该命名列的内容。
Here's the documentation for Views and Responses, which would help you with this issue:
这是 的文档Views and Responses，可以帮助您解决此问题：
回答by Neeraj Tangariya
You can simply use also this.
你也可以简单地使用这个。
public function($id){
$posts = Posts::where('uid','=',auth()->user()->id)->findOrFail($id);
return view('your_blade.php',compact('posts'));
OR either you can return this
return response()->json([$posts], 200);
}
Note: don't forgot to add trait also use App\Posts;
注意：不要忘记添加 trait 也使用 App\Posts；
