SQL 按天数分组

Question

我有兴趣查看从一些示例数据中获取包裹所需的不同长度。我目前正在使用此代码：

SELECT COUNT(*)
FROM orders_table
WHERE DATEDIFF(day, order_date, delivered_date) = 2
AND order_date BETWEEN '2021-01-01' AND '2021-01-31'

我想计算不同时长的送货数量，在上面的示例中，它查看了需要 2 天的送货数量。然而，这非常耗时，因为我需要在不同的天数内不断更改 where 语句并重新运行代码，是否有一种方法可以对其进行分组并在月份进行分组？这样输出看起来像这样：

January           February           March              April
2D   3D   4D      2D   3D   4D       2D   3D   4D      2D   3D   4D 
--------------------------------------------------------------------
12   7    32      21   53   33       8    22   41      9    44   30

Answer 1

这并没有完全按照您想要的格式提供输出，但结果似乎正是您想要的。通过 DATEDIFF 函数进行分组：

SELECT DATEDIFF(day, order_date, delivered_date) AS Date_Diff, COUNT(*)
FROM orders_table
WHERE order_date BETWEEN '2021-01-01' AND '2021-01-31'
GROUP BY DATEDIFF(day, order_date, delivered_date)

如果您只想要特定的 datediff 天数（例如，如果您只想查看订单和交货日期相差 2、3 和 4 天的订单），那么只需在过滤器中添加您想要的数字:

WHERE DATEDIFF(day, order_date, delivered_date) IN (2, 3, 4)

Answer 2

您的问题包括月份。为此：

SELECT DATE_TRUNC('month', order_date) as yyyymm
       DATEDIFF(day, order_date, delivered_date) as days_diff,
       COUNT(*)
FROM orders_table
GROUP BY DATE_TRUNC('month', order_date), DATEDIFF(day, order_date, delivered_date);

您可以使用条件聚合来透视这些结果：

SELECT DATEDIFF(day, order_date, delivered_date) as days_diff,
       SUM( (DATE_PART(month, order_date) = 1)::INT ) as jan,
       SUM( (DATE_PART(month, order_date) = 2)::INT ) as feb,
       SUM( (DATE_PART(month, order_date) = 3)::INT ) as mar,
       SUM( (DATE_PART(month, order_date) = 4)::INT ) as apr
       COUNT(*)
FROM orders_table
WHERE order_date >= '2021-01-01' AND
      order_date < '2021-05-01'
GROUP BY DATEDIFF(day, order_date, delivered_date);

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