2013-10-22 26 views
0

我想获取所有类的条目,其中有一些条件属性。一个关系,标签,应该急切加载。Laravel多对多急切加载总是返回空结果

的模型看起来像双外键,entry_id,TAG_ID所以

class Tag extends Eloquent { 
    protected $table = 'tags'; 
    protected $guarded = array(); 
    public function entries() 
    { 
     return $this->belongsToMany('Entry'); 
    } 
} 

class Entry extends Eloquent { 
    protected $table = 'entries'; 
    protected $guarded = array(); 
    public function tags() 
    { 
     return $this->belongsToMany('Tag'); 
    } 

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

    public function votes() 
    { 
     return $this->hasMany('Votes'); 
    } 
} 

表entry_tag,存在。

我正尝试使用此代码。 (1)

$testEntries = Entry::with(array('tags' => function($query) 
{ 
    $query->where('tag_id', '=', '1'); 
}))->get(); 

但是,它没有返回。 即使使用下面的代码也会产生完全的zilch。 (2)

$testEntries = Entry::with('tags')->get(); 

检查数据库日志,我可以看到,查询都OK。它们产生 (3)

select `tags`.*, `entry_tag`.`entry_id` as `pivot_entry_id`, `entry_tag`.`tag_id` as `pivot_tag_id` from `tags` inner join `entry_tag` on `tags`.`id` = `entry_tag`.`tag_id` where `entry_tag`.`entry_id` in (?, ?, ?, ?, ?, ?, ?, ?)","bindings":["1","2","3","4","5","6","7","8"] 

手动执行查询时(4)

select `tags`.*, `entry_tag`.`entry_id` as `pivot_entry_id`, `entry_tag`.`tag_id` as `pivot_tag_id` from `tags` inner join `entry_tag` on `tags`.`id` = `entry_tag`.`tag_id` where `entry_tag`.`entry_id` in (?, ?, ?, ?, ?, ?, ?, ?) and `tag_id` = ?","bindings":["1","2","3","4","5","6","7","8","1"] 

其中两个工作(并且发现结果)。

我错过了什么?我现在几个小时就挠了脑袋!

编辑:

我已经试过显示结果,没有任何运气,像下面

Log::debug('testing fetching entries:: ' . json_encode($testEntries)); 

foreach(Entry::with('tags')->get() as $entry) 
     { 
      Log::debug('test1!! ' . json_encode($entry));   
     } 

编辑2: 我曾尝试用他们的条目获取标签,如

Tag::with('entries')->get(); 

但它(连同其他组合)每次都返回零结果。我想也许我已经错过了我设置牌桌的方式。这里是完整的sql输出尝试(2),以防万一。

{"query":"select * from `entries`","bindings":[],"time":0.33},{"query":"select `tags`.*, `entry_tag`.`entry_id` as `pivot_entry_id`, `entry_tag`.`tag_id` as `pivot_tag_id` from `tags` inner join `entry_tag` on `tags`.`id` = `entry_tag`.`tag_id` where `entry_tag`.`entry_id` in (?, ?, ?, ?, ?, ?, ?, ?)","bindings":["1","2","3","4","5","6","7","8"],"time":0.74} 
+0

不确定这是否有帮助,但尝试定义您的BelongsToMany()关系,如下所示:'return $ this-> belongsToMany('Entry','entry_tag','entry_id','tag_id');'and vice在Entry表中输入 –

+0

你如何试图显示结果?也许在那里有一个错字。 – user1669496

+0

@GladToHelp,没有变化。由于它目前创建了成功的SQL查询(查看数据库日志),所以没有什么区别。 –

回答

0

如果使用软删除特质,检查deleted_at有默认值NULL