php - 我如何急切加载最新记录，其中每条记录都具有特定列的不同值？

Question

假设我有一张名为messages

的表

+----+--------------+-------------+-----+
| ID | user_id      | conv_id     |body |
+----+--------------+-------------+-----+
|1..5|            1 |           1 | ... |
+----+--------------+-------------+-----+
| 6  |            1 |           3 | ... |
+----+--------------+-------------+-----+
|  7 |            1 |           3 |...  |
+----+--------------+-------------+-----+
|  8 |            1 |           1 |...  |
+----+--------------+-------------+-----+
|  9 |            1 |           2 |...  |
+----+--------------+-------------+-----+
| 10 |            1 |           1 | ... |
+----+--------------+-------------+-----+
| 11 |            1 |           2 |...  |
+----+--------------+-------------+-----+ 
| 12 |            1 |           4 |...  |
+----+--------------+-------------+-----+
| 13 |            1 |           5 |...  |
+----+--------------+-------------+-----+
| 14 |            1 |           4 |...  |
+----+--------------+-------------+-----+

我想输出这个结果:

+----+--------------+-------------+-----+
|2..5|            1 |           1 | ... |
+----+--------------+-------------+-----+
| 10 |            1 |           1 | ... |
+----+--------------+-------------+-----+
|  7 |            1 |           3 | ... |
+----+--------------+-------------+-----+
| 11 |            1 |           2 | ... |
+----+--------------+-------------+-----+
| 13 |            1 |           5 |...  |
+----+--------------+-------------+-----+
| 14 |            1 |           4 |...  |
+----+--------------+-------------+-----+

作为预先加载的结果。如您所见，它输出 5 个最新记录，其中 conv_id = 1 加上每个其他 conv_id 的最新记录。我该怎么做？

---

这是我的 Controller 代码和我尝试过的代码

$loginuser = User::find(Auth::user()->id);

//$allOpenConvMsgs is an array of 5 message IDs where `conv_id = 1`.
//The relationship between user and messages is many-to-many. 
  See model codes below

$newMessages = $loginuser->messages()->where(function($q){
                   $q->where(''); //I'm stuck here!
               })->orWhereIn('messages.id',$allOpenConvMsgs);

消息模型

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

用户模型

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

Answer 1

如果我明白你的意思，你可以这样做:

$loginuser = User::find(Auth::user()->id)->with(array('messages' => function($q){
    $q->where('conv_id', '=', '1')->take(5)->orderBy('id', 'desc');
})->get();

$newMessages = $loginuser->messages;

更新

根据 David 的建议，最好使用通过 Auth::user() 函数获取的 User 实例。

所以上面的代码将是:

$newMessages = Auth::user()->load(array('messages' => function($q){
        $q->where('conv_id', '=', '1')->take(5)->orderBy('id', 'desc');
    }))->messages;

更新 2

上面的代码只获取conv_id = 1的消息，要获取所有最近的消息你必须移除where条件:

$newMessages = Auth::user()->load(array('messages' => function($q){
            $q->take(5)->orderBy('id', 'desc');
        }))->messages;