Question

我有一套Laravel设计系列：

"xdo45ttnqhsb" => array:5 [▼
      "design_id" => "xdo45ttnqhsb"
      "design_name" => "hufflepuff house"
      "category" => "harry potter"
      "tags" => array:6 [▶]
      "article_owner" => "1728"
    ]
    "wpy8r2erkk2t" => array:5 [▼
      "design_id" => "wpy8r2erkk2t"
      "design_name" => "ravenclaw house"
      "category" => "harry potter"
      "tags" => array:6 [▶]
      "article_owner" => "1728"
    ]
    "8oy7sb7i98q0" => array:5 [▼
      "design_id" => "8oy7sb7i98q0"
      "design_name" => "be positive friend"
      "category" => "miscelanea"
      "tags" => array:5 [▶]
      "article_owner" => "1728"
    ]
    "a9m79qc6bl9x" => array:5 [▼
      "design_id" => "a9m79qc6bl9x"
      "design_name" => "final fantasy pixel"
      "category" => "pixelart"
      "tags" => array:5 [▶]
      "article_owner" => "1728"
    ]

我想采用属于单一类别的设计。

例如：我只需要使用值为“harry potter”的“类别”设计。

我可以使用哪种收集助手？

Answer 1

为此，您可以使用适用于您的->filter()方法。

$collection->filter(function($arr){
    return $arr['category'] = "harry potter";
});

Answer 2

从laravel collections开始，根据我的知识，您可以使用filter()过滤记录，使用filter()进行编码，如下所示：

$collection->filter(function ($arrValue, $key) {
   return (isset($arrValue['category']) && $arrValue['category'] == "harry potter");
});

希望这有助于您解决问题。

Answer 3

您可以像其他人建议的那样使用filter()或使用：

$collection->where('category', '=', 'harry potter')->get()

这将返回上一个集合中category等于harry potter的项目集合。

在Laravel中过滤集合

3 个答案: