将var传递给雄辩的查询？

Question

public function getShow($id, $take, $skip)
{
    return $this->with(array('looks' => function($query){
        $query->wherePublished(1)->orderBy('sort')->take($take)->skip(0);
    }))->whereId($id)->wherePublished(1)->get(array('id', 'title', 'short_title', 'thumb', 'banner', 'video'));
}

我如何通过$ take？它说这是未定义的？

Answer 1

这样：

public function getShow($id, $take, $skip = 0)
{
    return $this->with(array('looks' => function($query) use ($take, $skip) 
    {
        $query->wherePublished(1)->orderBy('sort')->take($take)->skip($skip);
    }))->whereId($id)->wherePublished(1)->get(array('id', 'title', 'short_title', 'thumb', 'banner', 'video'));
}

Answer 2

使用scope进行方法链接和静态调用可能有用的另一种方法：

您的型号：

public function scopeGetShow($qry, $id, $take, $skip = 0)
{
    return $qry->with(array('looks' => function($query) use ($take, $skip) {
        $query->wherePublished(1)->orderBy('sort')->take($take)->skip($skip);
    }))->whereId($id)->wherePublished(1);
}

然后从控制器中使用它，如：

$result = ModelName::getShow()
         ->get(array('id', 'title', 'short_title', 'thumb', 'banner', 'video'));