php - 基于特定键值的唯一数组

Question

我有以下数组:

Array
(
    [0] => Array
    (
        [hotelID] => 10
        [hotelcategoryID] => 12
        [hotelName] => Grand Forest Metsovo
        [hotelShortDescription] => 
        [hotelVisible] => 1
        [roomID] => 2
    )

    [1] => Array
    (
        [hotelID] => 10
        [hotelcategoryID] => 12
        [hotelName] => Grand Forest Metsovo
        [hotelShortDescription] => 
        [hotelVisible] => 1
        [roomID] => 3
    )

    [2] => Array
    (
        [hotelID] => 10
        [hotelcategoryID] => 12
        [hotelName] => Grand Forest Metsovo
        [hotelShortDescription] => 
        [hotelVisible] => 1
        [roomID] => 4
    )

    [3] => Array
    (
        [hotelID] => 14
        [hotelcategoryID] => 7
        [hotelName] => Hotel Metropolis
        [hotelShortDescription] => 
        [hotelVisible] => 1
        [roomID] => 23
    )

    [4] => Array
    (
        [hotelID] => 14
        [hotelcategoryID] => 7
        [hotelName] => Hotel Metropolis
        [hotelShortDescription] => 
        [hotelVisible] => 1
        [roomID] => 24
    )

)

我有两个不同的酒店 ID key 。我只想提取一个元素(第一个元素)，其中 hotelID 在整个数组中是唯一的。我正在尝试使用以下代码:

$data['uniqueHotels'] = array_map('unserialize', array_unique(array_map('serialize', $hotels)));

但到目前为止还没有任何运气。

谁能给我一个提示？

Answer 1

如果寻找第一个元素:

<?php

$hotels = array(
  array(
    'id' => 1,
    'hotelID' => 10
  ),
  array(
    'id' => 2,
    'hotelID' => 10,
  ),
  array(
    'id' => 3,
    'hotelID' => 20,
  ),
  array(
    'id' => 4,
    'hotelID' => 20,
  ),
);


function getUniqueHotels($hotels) {
  $uniqueHotels = array();

  foreach($hotels as $hotel) {
    $niddle = $hotel['hotelID'];
    if(array_key_exists($niddle, $uniqueHotels)) continue;
    $uniqueHotels[$niddle] = $hotel;
  }

  return $uniqueHotels;
}

$unique_hotels = getUniqueHotels($hotels);
print_r($unique_hotels);

结果:

Array
(
    [10] => Array
        (
            [id] => 1
            [hotelID] => 10
        )

    [20] => Array
        (
            [id] => 3
            [hotelID] => 20
        )

)