Question

我正在尝试获取一个记录集，其中包含过去24小时内每小时的销售量，并用零填充空余时间。

我的示例数据集如下所示，并包含两个测试行：

id | saleID            | amount      | created
---|-------------------|-------------|-------------
1  | 6032b317-1533...  | 20000       | 2019-01-10 23:56:40
2  | 43556fg6-5344...  | 60000       | 2019-01-11 18:06:32

例如，当前时间是18:10，所以我要查找的结果是：

| hourOrderID | saleHour | saleHourTotal |
| ----------- | -------- | ------------- |
| 1           | 18:00    | 1             |
| 2           | 17:00    | 0             |
| 3           | 16:00    | 0             |
| 4           | 15:00    | 0             |
| 5           | 14:00    | 0             |
| 6           | 13:00    | 0             |
| 7           | 12:00    | 0             |
| 8           | 11:00    | 0             |
| 9           | 10:00    | 0             |
| 10          | 9:00     | 0             |
| 11          | 8:00     | 0             |
| 12          | 7:00     | 0             |
| 13          | 6:00     | 0             |
| 14          | 5:00     | 0             |
| 15          | 4:00     | 0             |
| 16          | 3:00     | 0             |
| 17          | 2:00     | 0             |
| 18          | 1:00     | 0             |
| 19          | 0:00     | 0             |
| 20          | 23:00    | 1             |
| 21          | 22:00    | 0             |
| 22          | 21:00    | 0             |
| 23          | 20:00    | 0             |
| 24          | 19:00    | 0             |
| 25          | 18:00    | 0             |

请注意；今天18：00-19：00之间有1笔交易，昨天18：00-19：00之间有0笔交易。这是我的问题。

我实际上得到的结果是：

| hourOrderID | saleHour | saleHourTotal |
| ----------- | -------- | ------------- |
| 1           | 18:00    | 1             |
| 2           | 17:00    | 0             |
| 3           | 16:00    | 0             |
| 4           | 15:00    | 0             |
| 5           | 14:00    | 0             |
| 6           | 13:00    | 0             |
| 7           | 12:00    | 0             |
| 8           | 11:00    | 0             |
| 9           | 10:00    | 0             |
| 10          | 9:00     | 0             |
| 11          | 8:00     | 0             |
| 12          | 7:00     | 0             |
| 13          | 6:00     | 0             |
| 14          | 5:00     | 0             |
| 15          | 4:00     | 0             |
| 16          | 3:00     | 0             |
| 17          | 2:00     | 0             |
| 18          | 1:00     | 0             |
| 19          | 0:00     | 0             |
| 20          | 23:00    | 1             |
| 21          | 22:00    | 0             |
| 22          | 21:00    | 0             |
| 23          | 20:00    | 0             |
| 24          | 19:00    | 0             |
| 25          | 18:00    | 1             | <----- this isn't yesterday, it's today

我一直在处理的查询是：

SELECT
    aux.id AS hourOrderID,
    CONCAT(aux.dh, ':00') AS saleHour,
    COALESCE(COUNT(ets.saleID), 0) AS saleHourTotal
FROM eventTicketSales AS ets 
RIGHT JOIN (
    SELECT 1 AS id, HOUR(UTC_TIMESTAMP()) AS dh UNION
    SELECT 2 AS id, HOUR(DATE_SUB(UTC_TIMESTAMP(), INTERVAL 1 HOUR)) AS dh UNION
    SELECT 3 AS id, HOUR(DATE_SUB(UTC_TIMESTAMP(), INTERVAL 2 HOUR)) AS dh UNION
    SELECT 4 AS id, HOUR(DATE_SUB(UTC_TIMESTAMP(), INTERVAL 3 HOUR)) AS dh UNION
    SELECT 5 AS id, HOUR(DATE_SUB(UTC_TIMESTAMP(), INTERVAL 4 HOUR)) AS dh UNION
    SELECT 6 AS id, HOUR(DATE_SUB(UTC_TIMESTAMP(), INTERVAL 5 HOUR)) AS dh UNION
    SELECT 7 AS id, HOUR(DATE_SUB(UTC_TIMESTAMP(), INTERVAL 6 HOUR)) AS dh UNION
    SELECT 8 AS id, HOUR(DATE_SUB(UTC_TIMESTAMP(), INTERVAL 7 HOUR)) AS dh UNION
    SELECT 9 AS id, HOUR(DATE_SUB(UTC_TIMESTAMP(), INTERVAL 8 HOUR)) AS dh UNION
    SELECT 10 AS id, HOUR(DATE_SUB(UTC_TIMESTAMP(), INTERVAL 9 HOUR)) AS dh UNION
    SELECT 11 AS id, HOUR(DATE_SUB(UTC_TIMESTAMP(), INTERVAL 10 HOUR)) AS dh UNION
    SELECT 12 AS id, HOUR(DATE_SUB(UTC_TIMESTAMP(), INTERVAL 11 HOUR)) AS dh UNION
    SELECT 13 AS id, HOUR(DATE_SUB(UTC_TIMESTAMP(), INTERVAL 12 HOUR)) AS dh UNION
    SELECT 14 AS id, HOUR(DATE_SUB(UTC_TIMESTAMP(), INTERVAL 13 HOUR)) AS dh UNION
    SELECT 15 AS id, HOUR(DATE_SUB(UTC_TIMESTAMP(), INTERVAL 14 HOUR)) AS dh UNION
    SELECT 16 AS id, HOUR(DATE_SUB(UTC_TIMESTAMP(), INTERVAL 15 HOUR)) AS dh UNION
    SELECT 17 AS id, HOUR(DATE_SUB(UTC_TIMESTAMP(), INTERVAL 16 HOUR)) AS dh UNION
    SELECT 18 AS id, HOUR(DATE_SUB(UTC_TIMESTAMP(), INTERVAL 17 HOUR)) AS dh UNION
    SELECT 19 AS id, HOUR(DATE_SUB(UTC_TIMESTAMP(), INTERVAL 18 HOUR)) AS dh UNION
    SELECT 20 AS id, HOUR(DATE_SUB(UTC_TIMESTAMP(), INTERVAL 19 HOUR)) AS dh UNION
    SELECT 21 AS id, HOUR(DATE_SUB(UTC_TIMESTAMP(), INTERVAL 20 HOUR)) AS dh UNION
    SELECT 22 AS id, HOUR(DATE_SUB(UTC_TIMESTAMP(), INTERVAL 21 HOUR)) AS dh UNION
    SELECT 23 AS id, HOUR(DATE_SUB(UTC_TIMESTAMP(), INTERVAL 22 HOUR)) AS dh UNION
    SELECT 24 AS id, HOUR(DATE_SUB(UTC_TIMESTAMP(), INTERVAL 23 HOUR)) AS dh UNION
    SELECT 25 AS id, HOUR(DATE_SUB(UTC_TIMESTAMP(), INTERVAL 24 HOUR)) AS dh
) AS aux ON HOUR(ets.created) = aux.dh
         AND ets.created > DATE_SUB(UTC_TIMESTAMP, INTERVAL 24 HOUR)
GROUP BY aux.id, aux.dh
ORDER BY hourOrderID

您可以fiddle with my bits here：o

我想要这样，所以集合中的最后一个小时是前一天，但是我的大脑决定它不能再走了，需要协助。

我尝试在第25小时添加另一个UNION，并尝试将时间范围延长到25小时，但结果与我想要的相去甚远：

AND ets.created > DATE_SUB(UTC_TIMESTAMP, INTERVAL 25 HOUR

请提出可以进行哪些更改以使其按预期工作。

Answer 1

我在注释部分采纳了@Strawberry的建议，并决定执行此应用程序级别，这现在很有意义，因为它更易于阅读。我不后悔尝试使用SQL版本，因为我学到了一点。

我在下面发布了我使用的解决方案，该解决方案在PHP中适用于需要类似功能的任何人。如果有人有比这更有效的解决方案，请分享评论。

查询：

SELECT
    CONCAT(DATE_FORMAT(created, '%Y-%m-%d %H'), ':00:00') AS saleHour,
    SUM(amount) AS totalSales
FROM eventTicketSales
WHERE created BETWEEN DATE_SUB(UTC_TIMESTAMP(), INTERVAL 24 HOUR) AND UTC_TIMESTAMP
GROUP BY HOUR(created);

结果：

saleHour                 totalSales
2019-01-11 17:00:00      10000
2019-01-10 23:00:00      20000

存储在$salesIn24Hours

中

逻辑：

$hours = 24 + 1;
for ($i=0; $i<$hours; $i++) {
    $date = date('Y-m-d H', strtotime('-'.$i.' HOUR')).':00:00';
    $key = array_search($date, array_column($salesIn24Hours, 'saleHour'));
    if ($key !== FALSE) {
        echo $date . ' ' . $salesIn24Hours[$key]['totalSales'] . '<br>';
    } else {
        echo $date . ' 0' . '<br>';
    }
}

结果：

2019-01-11 21:00:00  0
2019-01-11 20:00:00  0
2019-01-11 19:00:00  0
2019-01-11 18:00:00  0
2019-01-11 17:00:00  10000
2019-01-11 16:00:00  0
2019-01-11 15:00:00  0
2019-01-11 14:00:00  0
2019-01-11 13:00:00  0
2019-01-11 12:00:00  0
2019-01-11 11:00:00  0
2019-01-11 10:00:00  0
2019-01-11 09:00:00  0
2019-01-11 08:00:00  0
2019-01-11 07:00:00  0
2019-01-11 06:00:00  0
2019-01-11 05:00:00  0
2019-01-11 04:00:00  0
2019-01-11 03:00:00  0
2019-01-11 02:00:00  0
2019-01-11 01:00:00  0
2019-01-11 00:00:00  0
2019-01-10 23:00:00  20000
2019-01-10 22:00:00  0
2019-01-10 21:00:00  0

Answer 2

（出于语法目的，我假设您正在使用MySQL。）

为了从单个SELECT语句中产生25行数据，您需要一个具有25行的表（真实或伪造），这正是内部SELECT所做的。在显示实际构造内部表的更简单方法之前，我建议向其添加更多列（假设当前时间是12月31日下午6点）：

| hourOrderID | saleHour | startTime        | endTime          |
| ----------- | -------- | ---------------- | ---------------- |
| 1           | 18:00    | 2019-01-10 18:00 | 2019-01-10 19:00 |
| 2           | 17:00    | 2019-01-10 17:00 | 2019-01-10 18:00 |
| 3           | 16:00    | 2019-01-10 16:00 | 2019-01-10 17:00 |
| 4           | 15:00    | 2019-01-10 15:00 | 2019-01-10 16:00 |
| 5           | 14:00    | 2019-01-10 14:00 | 2019-01-10 15:00 |
| 6           | 13:00    | 2019-01-10 13:00 | 2019-01-10 14:00 |
| 7           | 12:00    | 2019-01-10 12:00 | 2019-01-10 13:00 |
| 8           | 11:00    | 2019-01-10 11:00 | 2019-01-10 12:00 |
| 9           | 10:00    | 2019-01-10 10:00 | 2019-01-10 11:00 |
| 10          | 9:00     | 2019-01-10 09:00 | 2019-01-10 10:00 |
| 11          | 8:00     | 2019-01-10 08:00 | 2019-01-10 09:00 |
| 12          | 7:00     | 2019-01-10 07:00 | 2019-01-10 08:00 |
| 13          | 6:00     | 2019-01-10 06:00 | 2019-01-10 07:00 |
| 14          | 5:00     | 2019-01-10 05:00 | 2019-01-10 06:00 |
| 15          | 4:00     | 2019-01-10 04:00 | 2019-01-10 05:00 |
| 16          | 3:00     | 2019-01-10 03:00 | 2019-01-10 04:00 |
| 17          | 2:00     | 2019-01-10 02:00 | 2019-01-10 03:00 |
| 18          | 1:00     | 2019-01-10 01:00 | 2019-01-10 02:00 |
| 19          | 0:00     | 2019-01-10 00:00 | 2019-01-10 01:00 |
| 20          | 23:00    | 2019-01-09 23:00 | 2019-01-10 00:00 |
| 21          | 22:00    | 2019-01-09 22:00 | 2019-01-09 23:00 |
| 22          | 21:00    | 2019-01-09 21:00 | 2019-01-09 22:00 |
| 23          | 20:00    | 2019-01-09 20:00 | 2019-01-09 21:00 |
| 24          | 19:00    | 2019-01-09 19:00 | 2019-01-09 20:00 |
| 25          | 18:00    | 2019-01-09 18:00 | 2019-01-09 19:00 |

然后您的查询将非常简单：

SELECT
    aux.hourOrderID,
    aux.saleHour,
    COALESCE(COUNT(ets.saleID), 0) AS saleHourTotal
  FROM madeUpTable AS aux LEFT JOIN eventTicketSales AS ets
      ON aux.startTime <= ets.created AND ets.created < aux.endTime
GROUP BY aux.hourOrderID
ORDER BY hourOrderID

事实证明，您可以使用通用表表达式（https://dev.mysql.com/doc/refman/8.0/en/with.html）来做到这一点：

WITH RECURSIVE aux AS
(
    SELECT 1 AS hourOrderId,
           HOUR(UTC_TIMESTAMP()) AS saleHour,
           DATE_ADD(
               DATE_FORMAT(UTC_TIMESTAMP(), "%Y-%m-%d %H:00:00"),
               INTERVAL 0 HOUR
           ) AS startDate,
           DATE_ADD(
               DATE_FORMAT(UTC_TIMESTAMP(), "%Y-%m-%d %H:00:00"),
               INTERVAL 1 HOUR
           ) AS endDate
    UNION ALL
    SELECT n + 1,
           HOUR(DATE_SUB(startDate, INTERVAL 1 HOUR)),
           DATE_SUB(startDate, INTERVAL 1 HOUR),
           DATE_SUB(startDate, INTERVAL 1 HOUR)
      FROM aux
     WHERE n < 25
)
SELECT
    aux.hourOrderID,
    aux.saleHour,
    COALESCE(COUNT(ets.saleID), 0) AS saleHourTotal
  FROM madeUpTable AS aux LEFT JOIN eventTicketSales AS ets
      ON aux.startTime <= ets.created AND ets.created < aux.endTime
GROUP BY aux.hourOrderID
ORDER BY hourOrderID

我个人试图限制这种技术在涉及其他人的代码上的使用，因为这不是大多数人在野外看到的东西，但是对于SQL来说，这对我来说更容易理解，因为查询本身很简单。恕我直言，其中最复杂的部分实际上是将当前时间四舍五入到最近的小时，同时保留日期部分。

最近24小时内每小时的返回计数，用零填充空白

2 个答案: