我正在尝试构建一组Eloquent模型,它们代表现有的硬件设备数据库(两者都不能更改).我知道如何在sql中执行此操作,但我正在努力构建一个使用第三个表的模型关系,类似于关系/联结表,但是与复合键一对一的关系.
有三个实体(简化):
>设备
>会议
> device_user
用户可以同时在许多设备中,并且具有与这些设备相关联的会话日志.用户确实拥有唯一的ID,但从设备的角度来看,它们只有一个“用户号”,它只是短(3个字节),因此不能代表整个用户范围,因此它映射在device_user表中. (它实际上比这更复杂,但出于这个问题的目的,我已经将其剥离了)
设备表:
d_id PK [data fields...]
device_user表:
du_uid User's actual ID du_device_id FK to device.d_id du_number 000-999 [Metadata...]
会话表:
s_device_id device.d_id s_user_number 000-999 (device_user.du_number) [data fields...]
场景:我有一个会话,我想查找特定的device_user.d_uid.在sql中我会做类似的事情:
SELECT session.blah,du_uid FROM session INNER JOIN device_user ON du_device_id = s_device_id AND du_number = s_user_number
所以我想这实际上只是一个复合键的关系.
我在Eloquent中尝试的是这样的:
class SessionLogModel { public function device(){ return $this->belongsTo('MyModels\\DeviceModel','s_device_id','d_id'); } public function user(){ return $this->belongsTo('MyModels\\DeviceUserModel','s_user_number','du_number') // A) I tried: ->withDevice($this->s_device_id); // or B) I tried: ->withDevice($this->device()); } // example usage public static function getRecentUser(DateTime $localTime,$deviceId){ $u = null; // get the preceding session log $q = SessionLogModel::where('session_type','=','S') ->where('session_device_id',$deviceId) ->where('sesison_date','<=',$localTime) ->orderBy('session_id','DESC') ->take(1) ->with('device') ->with('user'); $s = $q->get()->first(); $u = $s->user->du_uid; // use the unique user ID ... } } class DeviceUserModel { // A) public function scopeWithDevice($query,$device_id){ return $query->where('du_device_id',$device_id); } // OR B) I tried: public function scopeWithDevice($query,$device){ return $query->where('du_device_id',$device->d_id); } }
我已经尝试了很多方法来限制匹配两个列的范围或其他’where’结构,但似乎总是很难通过BelongsTo“发送”正确的值.在检查DB :: getQueryLog时,设备ID以NULL形式出现.但是,如果我对属性中的值进行硬编码,我可以看到它“正常工作”.
我对此进行了相当多的研究,但我发现很难找到类似的结构.
我使用Laravel v4.2中的Eloquent独立使用(不在Laravel中).
我刚刚遇到了这个有趣的问题:
我厌倦了如下所示在laravel中模拟你的桌子如下:
我厌倦了如下所示在laravel中模拟你的桌子如下:
public function up(){ Schema::create('session',function (Blueprint $table) { $table->increments('id'); $table->integer('s_device_id'); $table->string('s_user_number',20); $table->timestamps(); }); Schema::create('device',function (Blueprint $table) { $table->increments('d_id'); $table->string('blah',20); $table->timestamps(); }); Schema::create('device_user',function (Blueprint $table) { $table->integer('du_device_id')->unsigned(); $table->integer('du_uid')->unsigned(); $table->string('du_number',20); $table->primary(['du_device_id','du_uid']);//important composite key $table->timestamps(); }); } //then do the relations on Medels: //User Model public function deviceUser(){ return $this->hasOne(DeviceUser::class,'du_uid'); } //Device Model public function deviceUser(){ return $this->hasOne(DeviceUser::class,'du_device_id','d_id'); } //DeviceUser Model public function device(){ return $this->belongsTo(Device::class,'d_id'); } public function user(){ return $this->belongsTo(User::class,'du_uid'); } //Session Model //[Not the Session Facade of Laravel] public function device(){ return $this->belongsTo(Device::class,'s_device_id'); } //Now let us do the work in SessionController after filling your tables with demo data for e.g. //all these relations are working fine! $device = Device::where('d_id',1)->first(); $user = User::where('id',4)->first(); //dd($user,$user->deviceUser,$device,$device->deviceUser);//here instance objects and their relations can be fetched easily $device_user = DeviceUser::where('du_device_id',1)->where('du_uid',4)->first(); //dd($device_user,$device_user->device); //$session = Session::where('id',100)->first();//can get session by ID $session = Session::where('s_device_id',1)->where('s_user_number','000-999')->first();//get session by unique composite key which what you are after. It is similar to the sql Query that you built. Then you can easily fetch the relations as follows: dd($session,$session->device,$session->device->deviceUser);
希望这会有所帮助!