SQL计数按天数分组

Question

我感兴趣的是，从一些示例数据中获取一个包需要不同的长度。我目前使用的是以下代码：

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

我想要得到需要不同长度的交付计数，在上面的例子中，它查看了需要2天的交付数量。然而，这相当耗时，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，是否有方法对其进行排序并按月进行分组？因此，输出将如下所示：

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

这并不能准确地提供所需格式的输出，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，如果您想查看订单和交货日期只有2天、3天和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。要做到这一点：

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);