Question

（见下面的代码）

我有两个模型User和Messages。我想根据每个输入的会话ID获取10条消息 - 及其相关的发件人姓名，并反转数据。我已经成功获取了消息，但无法将其缩小为每个会话/其相关发件人姓名的10条消息。

我试过用这些用户数据返回消息数据但是我不知道如何访问发件人的名字，或者它没有返回关联的用户：
```
$allMessages = Messages::with('User')->whereIn('conv_id',$conv_id);
```
我也试过这个回复10条消息，但它根据收到的所有消息而不是基于每个对话返回10条消息。
```
$allMessages = Messages::whereIn('conv_id',$conv_id)->take(10);
```

如何根据每个会话的ID以及消息的相关发件人名称获取10条消息？改善代码的奖励点。提前谢谢！

用户模型

public function messages(){
    return $this->hasMany('Messages');
}

消息模型

public function user(){
    return $this->belongsTo('User');
}

控制器

public function index()
    {
        $timestamps = Input::get('from'); //timestamp of latest message
        $conv_id = Input::get('conv_id'); //array of conversation ids
        $allMessages = Messages::whereIn('conv_id',$conv_id);
        if(is_null($timestamps)){
           $messages = $allMessages->orderBy('created_at','desc');
        }else{
           asort($timestamps);
           $messages = $allMessages->where('created_at','>',end($timestamps));
        }
        return $messages->get()->reverse();
    }

迁移

消息表

Schema::create('messages', function(Blueprint $table)
        {
            $table->increments('id');
            $table->integer('user_id');
            $table->integer('conv_id');
            $table->text('body');
            $table->timestamps();
        });

用户表

Schema::create('users',function($table)
        {
            $table->increments('id');
            $table->string('email')->unique();
            $table->string('password',100);
            $table->string('name',150);
            $table->string('usertype',50);
            $table->boolean('block');
            $table->string('remember_token',100);
            $table->timestamp('lastlogin_at');
            $table->timestamps();
            $table->softDeletes();
        });

编辑：var_dump WereWolf在Chrome中回答的预览结果

0: {id:37, user_id:1, conv_id:2, body:damn mate, created_at:2014-06-20 00:55:32,…}
1: {id:38, user_id:1, conv_id:2, body:hello, created_at:2014-06-20 02:18:21,…}
2: {id:39, user_id:1, conv_id:2, body:dude, created_at:2014-06-20 02:20:10,…}
3: {id:40, user_id:1, conv_id:2, body:test, created_at:2014-06-20 06:52:37,…}
4: {id:67, user_id:1, conv_id:2, body:haha, created_at:2014-06-21 01:25:56,…}
5: {id:68, user_id:1, conv_id:2, body:hey, created_at:2014-06-21 01:26:14, updated_at:2014-06-21 01:26:14,…}
6: {id:69, user_id:1, conv_id:1, body:testa, created_at:2014-06-21 01:27:02,…}
7: {id:70, user_id:1, conv_id:1, body:testser, created_at:2014-06-21 01:27:32,…}
8: {id:115, user_id:1, conv_id:2, body:haha, created_at:2014-06-21 02:45:06,…}
9: {id:116, user_id:1, conv_id:2, body:test, created_at:2014-06-21 02:57:03,…}

Answer 1

你可以试试这个：

$allMessages = Messages::with('user')
                       ->where('conv_id', $conv_id)
                       ->take(10)
                       ->get();

这将返回Messages模型的集合，因此您需要循环：

@foreach($allMessages as $message)
    {{ $message->user->name }}
    {{ $message->body }}
@endforeach

另外，您可以使用：

// Get the first message and then user->name
$allMessages->first()->user->name;

// Get the second message and then user->name
$allMessages->get(1)->user->name;

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