Question

我正在使用名为appointments的表，这些列是：

scheduled_at：日期时间 doctor_id：整数说明：文字 finish_at：datetime

我想获得按doctor_id分组以及按一组日期间隔分组的所有约会的计数，这些日期间隔根据给定的输入集而变化。我得到的输入是：from是日期，to是日期。

我要解决的第一个用例是将给定的from和to中的结果按月划分。因此，如果给我from ='2018-02-03'和to ='2018-10-03'，我想获取为每个医生完成的所有约会的COUNT，并且从2018-02到2018-10的每个月。

我知道如何按Doctor_id进行分组，但是我不知道如何对这些动态日期间隔进行分组。

这是我到目前为止所拥有的：

SELECT COUNT(*) FROM appointments WHERE finished_at IS NOT NULL GROUP BY doctor_id

我知道，如果事先有间隔，我可以进行尽可能多的查询，因此最终我每个月都要进行一次查询，例如：

SELECT COUNT(*) FROM appointments WHERE finished_at IS NOT NULL AND scheduled_at BETWEEN '2018-02-01' AND '2018-03-01' GROUP BY doctor_id

SELECT COUNT(*) FROM appointments WHERE finished_at IS NOT NULL AND scheduled_at BETWEEN '2018-03-01' AND '2018-04-01' GROUP BY doctor_id

我只是想知道是否有一种方法可以在SQL中做到这一点，所以我只需要执行一个查询。

Answer 1

您可以使用条件聚合：

SELECT doctor_id,
       SUM(CASE WHEN scheduled_at BETWEEN '2018-02-01' AND '2018-03-01' THEN 1 ELSE 0 END),
       SUM(CASE WHEN scheduled_at BETWEEN '2018-03-01' AND '2018-04-01' THEN 1 ELSE 0 END)    
FROM appointments
WHERE finished_at IS NOT NULL
GROUP BY doctor_id

Answer 2

使用case表达式进行条件聚合：

SELECT doctor_id,
       COUNT(*),
       COUNT(case when scheduled_at BETWEEN '2018-02-01' AND '2018-03-01' then 1 end),
       COUNT(case when scheduled_at BETWEEN '2018-03-01' AND '2018-04-01' then 1 end)
FROM appointments
WHERE finished_at IS NOT NULL
GROUP BY doctor_id

或者也有group by年和月的查询：

SELECT doctor_id, year(scheduled_at), month(scheduled_at), COUNT(*)
FROM appointments
WHERE finished_at IS NOT NULL
GROUP BY doctor_id, year(scheduled_at), month(scheduled_at)

或者，也许应该使用ANSI SQL函数EXTRACT：

SELECT doctor_id,
       extract(year from scheduled_at), extract(month from scheduled_at), COUNT(*)
FROM appointments
WHERE finished_at IS NOT NULL
GROUP BY doctor_id, extract(year from scheduled_at), extract(month from scheduled_at)

Answer 3

如果您的postgres版本是9.4或更高版本，则应考虑使用FILTER谓词，它比case语句更易读，性能更高。

SELECT doctor_id,
       COUNT(*) AS total,
       COUNT(*) FILTER (WHERE scheduled_at BETWEEN '2018-02-01' AND '2018-03-01') AS date_feb,
       COUNT(*) FILTER (WHERE scheduled_at BETWEEN '2018-03-01' AND '2018-04-01') AS date_mar
FROM appointments
WHERE finished_at IS NOT NULL
GROUP BY doctor_id

GROUP BY外键和日期范围

3 个答案: