根據同一數組中的另一個唯一值將一組值推入數組

Question

問題背景

您好，我有以下電影攝制組成員：

array:7 [▼
  0 => array:6 [▼
    "credit_id" => "52fe49dd9251416c750d5e9d"
    "department" => "Directing"
    "id" => 139098
    "job" => "Director"
    "name" => "Derek Cianfrance"
    "profile_path" => "/zGhozVaRDCU5Tpu026X0al2lQN3.jpg"
  ]
  1 => array:6 [▼
    "credit_id" => "52fe49dd9251416c750d5ed7"
    "department" => "Writing"
    "id" => 139098
    "job" => "Story"
    "name" => "Derek Cianfrance"
    "profile_path" => "/zGhozVaRDCU5Tpu026X0al2lQN3.jpg"
  ]
  2 => array:6 [▼
    "credit_id" => "52fe49dd9251416c750d5edd"
    "department" => "Writing"
    "id" => 132973
    "job" => "Story"
    "name" => "Ben Coccio"
    "profile_path" => null
  ]
  3 => array:6 [▼
    "credit_id" => "52fe49dd9251416c750d5ee3"
    "department" => "Writing"
    "id" => 139098
    "job" => "Screenplay"
    "name" => "Derek Cianfrance"
    "profile_path" => "/zGhozVaRDCU5Tpu026X0al2lQN3.jpg"
  ]
  4 => array:6 [▼
    "credit_id" => "52fe49dd9251416c750d5ee9"
    "department" => "Writing"
    "id" => 132973
    "job" => "Screenplay"
    "name" => "Ben Coccio"
    "profile_path" => null
  ]
  5 => array:6 [▼
    "credit_id" => "52fe49dd9251416c750d5eef"
    "department" => "Writing"
    "id" => 1076793
    "job" => "Screenplay"
    "name" => "Darius Marder"
    "profile_path" => null
  ]
  11 => array:6 [▼
    "credit_id" => "52fe49de9251416c750d5f13"
    "department" => "Camera"
    "id" => 54926
    "job" => "Director of Photography"
    "name" => "Sean Bobbitt"
    "profile_path" => null
  ]
]

如您所見，這是我通過TMDb API獲得的積分列表。 構建上述數組的第一步是過濾掉我不想顯示的所有作業，這是我的操作方法：

$jobs = [ 'Director', 'Director of Photography', 'Cinematography', 'Cinematographer', 'Story', 'Short Story', 'Screenplay', 'Writer' ];

$crew = array_filter($tmdbApi, function ($crew) use ($jobs) {
    return array_intersect($jobs, $crew);
});

我的問題

我想弄清楚如何使上述結果更進一步，並合並id相同的jobs ，從而得到這樣的結果，例如：

array:7 [▼
  0 => array:6 [▼
    "credit_id" => "52fe49dd9251416c750d5e9d"
    "department" => "Directing"
    "id" => 139098
    "job" => "Director, Story, Screenplay"
    "name" => "Derek Cianfrance"
    "profile_path" => "/zGhozVaRDCU5Tpu026X0al2lQN3.jpg"
  ]

我也考慮過按照我的邏輯去做，而不是在我的刀片模板中去做，但是我不確定如何實現。

您將如何實現？

Answer 1

由於您正在嘗試編輯數組元素及其大小，因此我相信array_map()或array_filter()將不是解決方案。

這就是我能想到的...

$jobs = [
  'Director', 'Director of Photography', 'Cinematography',
  'Cinematographer', 'Story', 'Short Story', 'Screenplay', 'Writer'
];

$crew = [];

foreach($tmdbApi as $key => $member) {
  if($member['id'] == $id && in_array($member['job'], $jobs)) {
    if(!isset($crew[$key]))  {
      $crew[$key] = $member;
    } else {
      $crew_jobs = explode(', ', $crew[$key]['job']);
      if(!in_array($member['job'], $crew_jobs)) {
        $crew_jobs[] = $member['job'];
      }
      $crew[$key]['job'] = implode(', ', $crew_jobs);
    }
  }
}

希望這能回答您的問題:)

Answer 2

在這種情況下，您可以很好地使用Laravel的Collection，它有很多方法可以在這種情況下為您提供幫助。

首先，將此數組（您已經在作業中過濾的數組）轉到集合：

$collection = collect($crew);

其次，按id這個Collection分組：

$collectionById = $collection->groupBy('id');

現在，將結果按 id 分組，並轉換為一個集合，其中鍵對應於id，其值是一個“匹配”結果數組。 有關此的更多信息。

最后，只是一個簡單的腳本，它遍歷每個id所有結果並組合job字段：

$combinedJobCollection = $collectionById->map(function($item) {
    // get the default object, in which all fields match
    // all the other fields with same ID, except for 'job'
    $transformedItem = $item->first();

    // set the 'job' field according all the (unique) job
    // values of this item, and implode with ', '
    $transformedItem['job'] = $item->unique('job')->implode('job', ', ');

    /* or, keep the jobs as an array, so blade can figure out how to output these
    $transformedItem['job'] = $item->unique('job')->pluck('job');
    */

    return $transformedItem;
})->values();
// values() makes sure keys are reordered (as groupBy sets the id
// as the key)

此時，將返回此Collection：

Collection {#151 ▼
  #items: array:4 [▼
    0 => array:6 [▼
      "credit_id" => "52fe49dd9251416c750d5e9d"
      "department" => "Directing"
      "id" => 139098
      "job" => "Director, Story, Screenplay"
      "name" => "Derek Cianfrance"
      "profile_path" => "/zGhozVaRDCU5Tpu026X0al2lQN3.jpg"
    ]
    1 => array:6 [▼
      "credit_id" => "52fe49dd9251416c750d5edd"
      "department" => "Writing"
      "id" => 132973
      "job" => "Story, Screenplay"
      "name" => "Ben Coccio"
      "profile_path" => null
    ]
    2 => array:6 [▼
      "credit_id" => "52fe49dd9251416c750d5eef"
      "department" => "Writing"
      "id" => 1076793
      "job" => "Screenplay"
      "name" => "Darius Marder"
      "profile_path" => null
    ]
    3 => array:6 [▼
      "credit_id" => "52fe49de9251416c750d5f13"
      "department" => "Camera"
      "id" => 54926
      "job" => "Director of Photography"
      "name" => "Sean Bobbitt"
      "profile_path" => null
    ]
  ]
}

注意：將此Collection用作數組，請使用：

$crew = $combinedJobCollection->toArray();

有多種方法可以實現此目的，例如：在數組中search重疊的ID，但是我認為這是實現此目的的最簡單方法。

祝好運！