按一个深列对 3d 数组中的条目进行分组，并对每组另一个深列求和

Question

如何按成人价格分组，然后获得每组成人人数的总和。

输入示例：

[
    [
        'Reservation' => ['adult_price' => 2000, 'adult_count' => 2],
    ],
    [
        'Reservation' => ['adult_price' => 2000, 'adult_count' => 1],
    ],
    [
        'Reservation' => ['adult_price' => 300, 'adult_count' => 1],
    ],
]

预期结果：

[
    [
        'Reservation' => ['adult_price' => 2000, 'adult_count' => 3],
    ],
    [
        'Reservation' => ['adult_price' => 300, 'adult_count' => 1],
    ],
]

Answer 1

试试这个：

$grouped = [];

foreach ($array as $item) {
    $item = $item["Reservation"];
    $currentValue = isset($grouped[$item["adult_price"]]) ? $grouped[$item["adult_price"]] : 0;
    $grouped[$item["adult_price"]] = $currentValue + $item["adult_count"];
}

$buildArray = [];

foreach ($grouped as $price => $count) {
    $buildArray[] = ["Reservation" => ["adult_price" => $price, "adult_count" => $count]];
}

print_r($buildArray);

https://3v4l.org/iZZ4m

Answer 2

尝试这样的事情：

function summarize($ReservationArr){
  $summed = [];
  foreach ($ReservationArr as $one){

    $adult_price = $one["Reservation"]["adult_price"];
    $adult_count= $one["Reservation"]["adult_count"];
    if (isset( $summed[$adult_price])){
            $summed[$adult_price] += (int)$adult_count;
    } else {
          $summed[$adult_price] = (int)$adult_count;
        }
  }
    // now rebuild:
    $result = [];
    foreach ($summed as $price => $number){
        $result[] = array("Reservation" => array("adult_price" => $price, "adult_count" => $number));
    }
    return $result;
}

$test = [];
$test[] = array("Reservation" => array("adult_price" => 2000, "adult_count" => 2));
$test[] = array("Reservation" => array("adult_price" => 1000, "adult_count" => 1));
$test[] = array("Reservation" => array("adult_price" => 500, "adult_count" => 4));
$test[] = array("Reservation" => array("adult_price" => 2000, "adult_count" => 3));

$test[] = array("Reservation" => array("adult_price" => 2000, "adult_count" => 2));
$test[] = array("Reservation" => array("adult_price" => 1000, "adult_count" => 1));
$test[] = array("Reservation" => array("adult_price" => 500, "adult_count" => 4));
$test[] = array("Reservation" => array("adult_price" => 2000, "adult_count" => 3));

$res = summarize($test);
echo "<pre>";
print_r($res);
echo "</pre>";

Answer 3

使用两个循环是太多不必要的工作。将输出数组视为持续更新的查找数组。这将确定您正在处理的是新的 Adult_price 还是以前遇到过的。使用

$row['Reservation']['adult_price']

值作为新的临时第一级键 - 这有助于

isset()

了解是否应保存整行，或者是否应将 Adult_count 添加到存储的计数中。

代码：（演示）

$rows = [
    ["Reservation" => ["adult_price" => 2000, "adult_count" => 2]],
    ["Reservation" => ["adult_price" => 1000, "adult_count" => 1]],
    ["Reservation" => ["adult_price" => 500, "adult_count" => 4]],
    ["Reservation" => ["adult_price" => 2000, "adult_count" => 3]],
    ["Reservation" => ["adult_price" => 2000, "adult_count" => 2]],
    ["Reservation" => ["adult_price" => 1000, "adult_count" => 1]],
    ["Reservation" => ["adult_price" => 500, "adult_count" => 4]],
    ["Reservation" => ["adult_price" => 2000, "adult_count" => 3]],
];

$result = [];
foreach ($rows as $row) {
    if (!isset($result[$row['Reservation']['adult_price']])) {
        $result[$row['Reservation']['adult_price']] = $row;
    } else {
        $result[$row['Reservation']['adult_price']]['Reservation']['adult_count'] += $row['Reservation']['adult_count'];
    }
}
var_export(array_values($result));

或者，这可能会减少眼睛疲劳：

foreach ($rows as $row) {
    $key = $row['Reservation']['adult_price'];
    if (!isset($result[$key])) {
        $result[$key] = $row;
    } else {
        $result[$key]['Reservation']['adult_count'] += $row['Reservation']['adult_count'];
    }
}

或者将行数据解构为方便的变量，并将唯一行作为引用推入结果数组中，然后仅添加新计数来存储重新遇到价格组时的计数。

代码：（演示）

$result = [];
foreach ($rows as $row) {
    ['Reservation' => ['adult_price' => $p, 'adult_count' => $c]] = $row;
    if (!isset($ref[$p])) {
        $ref[$p] = $row;
        $result[] =& $ref[$p];
        continue;
    }
    $ref[$p]['Reservation']['adult_count'] += $c;
}
var_export($result);

输出：（来自任一片段）

array (
  0 => 
  array (
    'Reservation' => 
    array (
      'adult_price' => 2000,
      'adult_count' => 10,
    ),
  ),
  1 => 
  array (
    'Reservation' => 
    array (
      'adult_price' => 1000,
      'adult_count' => 2,
    ),
  ),
  2 => 
  array (
    'Reservation' => 
    array (
      'adult_price' => 500,
      'adult_count' => 8,
    ),
  ),
)