PHP实现：按连续日期与相同MinStay值分组日期二维数组

我有2023-11-01至2024-01-04的日期数据，部分日期对应不同的MinStay值。现有输入数组：

$input = [
    ['date' => '2023-11-01', 'MinStay' => 1],
    ['date' => '2023-11-02', 'MinStay' => 1],
    ['date' => '2023-11-03', 'MinStay' => 1],
    ['date' => '2023-11-04', 'MinStay' => 2],
    ['date' => '2023-11-05', 'MinStay' => 2],
    ['date' => '2023-11-06', 'MinStay' => 2],
    ['date' => '2023-12-10', 'MinStay' => 1],
    ['date' => '2023-12-11', 'MinStay' => 1],
    ['date' => '2023-12-12', 'MinStay' => 3],
    ['date' => '2023-12-13', 'MinStay' => 2],
    ['date' => '2023-12-14', 'MinStay' => 2],
    ['date' => '2024-01-01', 'MinStay' => 4],
    ['date' => '2024-01-02', 'MinStay' => 4],
    ['date' => '2024-01-03', 'MinStay' => 4],
    ['date' => '2024-01-04', 'MinStay' => 4],
];

期望将连续日期且MinStay相同的条目合并为日期范围，得到如下输出：

$output = [
    ['dateFrom' => '2023-11-01', 'dateTo' => '2023-11-03', 'MinStay' => 1],
    ['dateFrom' => '2023-11-04', 'dateTo' => '2023-11-06', 'MinStay' => 2],
    ['dateFrom' => '2023-12-10', 'dateTo' => '2023-12-11', 'MinStay' => 1],
    ['dateFrom' => '2023-12-12', 'dateTo' => '2023-12-12', 'MinStay' => 3],
    ['dateFrom' => '2023-12-13', 'dateTo' => '2023-12-14', 'MinStay' => 2],
    ['dateFrom' => '2024-01-01', 'dateTo' => '2024-01-04', 'MinStay' => 4],
];

我尝试了以下代码，但未达到预期效果：

$arr = [];
foreach ($input as $date) {
    if ($date['MinStay'] == 1) {
        $arr[] = [
            'dateFrom' => $date['date'],
            'dateTo' => $date['date'],
            'MinStay' => $date['MinStay'],
        ];
    }
}

解决方案 #

核心思路是遍历过程中跟踪当前合并的分组，判断当前条目是否和当前分组满足「MinStay相同且日期连续」的条件，满足则更新分组结束日期，否则存入当前分组并创建新分组。

完整实现代码：

$input = [
    ['date' => '2023-11-01', 'MinStay' => 1],
    ['date' => '2023-11-02', 'MinStay' => 1],
    ['date' => '2023-11-03', 'MinStay' => 1],
    ['date' => '2023-11-04', 'MinStay' => 2],
    ['date' => '2023-11-05', 'MinStay' => 2],
    ['date' => '2023-11-06', 'MinStay' => 2],
    ['date' => '2023-12-10', 'MinStay' => 1],
    ['date' => '2023-12-11', 'MinStay' => 1],
    ['date' => '2023-12-12', 'MinStay' => 3],
    ['date' => '2023-12-13', 'MinStay' => 2],
    ['date' => '2023-12-14', 'MinStay' => 2],
    ['date' => '2024-01-01', 'MinStay' => 4],
    ['date' => '2024-01-02', 'MinStay' => 4],
    ['date' => '2024-01-03', 'MinStay' => 4],
    ['date' => '2024-01-04', 'MinStay' => 4],
];

$output = [];
$currentGroup = null;

foreach ($input as $item) {
    $currentDate = new DateTime($item['date']);
    
    if ($currentGroup === null) {
        $currentGroup = [
            'dateFrom' => $item['date'],
            'dateTo' => $item['date'],
            'MinStay' => $item['MinStay']
        ];
        continue;
    }
    
    $groupEndDate = new DateTime($currentGroup['dateTo']);
    $nextDay = (clone $groupEndDate)->modify('+1 day');
    
    // 判断是否属于同一分组：MinStay相同且日期是上一日期的次日
    if ($item['MinStay'] === $currentGroup['MinStay'] && $currentDate == $nextDay) {
        $currentGroup['dateTo'] = $item['date'];
    } else {
        $output[] = $currentGroup;
        $currentGroup = [
            'dateFrom' => $item['date'],
            'dateTo' => $item['date'],
            'MinStay' => $item['MinStay']
        ];
    }
}

// 加入最后一个分组
if ($currentGroup !== null) {
    $output[] = $currentGroup;
}

// 验证结果
print_r($output);

代码关键点

• 使用DateTime对象处理日期，避免手动解析字符串的误差，准确判断日期连续性
• 通过currentGroup变量动态维护当前合并的分组，无需提前分组
• 遍历结束后必须将最后一个未存入的分组加入结果数组

内容的提问来源于stack exchange，提问作者Hola