SQL - 有没有办法计算一系列日期的运行总数？

Question

给定一个带有日期的事件列表，其中 1 表示某个人（例如学生）加入了一个小组，-1 表示某个人离开了一个小组，是否可以在 SQL 中按日期计算小组规模？我有代码可以生成一个范围内的所有日期......当我运行它自己工作时。然后我想在 class 之前参加注册活动，并获得每个日期注册的总人数。（+1 == class 添加，-1 == class 删除）。

我想我遗漏了一些关于连接和分组在 SQL 中如何工作的基本知识。

http://sqlfiddle.com/#!9/e4835/5/0

示例数据：

CREATE TABLE classes(`id` int, `name` varchar(7));
INSERT INTO classes(`id`, `name`) VALUES
    (1, 'math'),
    (2, 'english'),
    (3, 'sciene');

CREATE TABLE enrollment_changes(
  `class_id` int, 
  `change_date` date, 
  `change` int);
INSERT INTO enrollment_changes
    (`class_id`, `change_date`, `change`)
VALUES
    (1, '2019-01-01', 1),
    (1, '2019-01-01', 1),
    (1, '2019-01-02', -1),
    (3, '2019-01-02', 1),
    (1, '2019-01-03', 1),
    (2, '2019-01-03', -1)
;

-- This gets me part way there... it produces the product of dates x classes 
SELECT 
  date_range.event_date, c.name
FROM 
  (SELECT adddate('1970-01-01',t4*10000 + t3*1000 + t2*100 + t1*10 + t0) event_date 
    FROM
      (SELECT 0 t0 UNION SELECT 1 UNION SELECT 2 UNION SELECT 3 UNION SELECT 4 UNION SELECT 5 UNION SELECT 6 UNION SELECT 7 UNION SELECT 8 UNION SELECT 9) t0,
      (SELECT 0 t1 UNION SELECT 1 UNION SELECT 2 UNION SELECT 3 UNION SELECT 4 UNION SELECT 5 UNION SELECT 6 UNION SELECT 7 UNION SELECT 8 UNION SELECT 9) t1,
      (SELECT 0 t2 UNION SELECT 1 UNION SELECT 2 UNION SELECT 3 UNION SELECT 4 UNION SELECT 5 UNION SELECT 6 UNION SELECT 7 UNION SELECT 8 UNION SELECT 9) t2,
      (SELECT 0 t3 UNION SELECT 1 UNION SELECT 2 UNION SELECT 3 UNION SELECT 4 UNION SELECT 5 UNION SELECT 6 UNION SELECT 7 UNION SELECT 8 UNION SELECT 9) t3,
      (SELECT 0 t4 UNION SELECT 1 UNION SELECT 2 UNION SELECT 3 UNION SELECT 4 UNION SELECT 5 UNION SELECT 6 UNION SELECT 7 UNION SELECT 8 UNION SELECT 9) t4
  ) AS date_range
  JOIN 
    classes c
WHERE 
  date_range.event_date BETWEEN '2019-01-01' AND '2019-01-03'
;

-- This does not work at all... it reduces the output to a single record.
SELECT 
  date_range.event_date, c.name, SUM(e.change) AS 'NetEnrollment'
FROM 
  (SELECT adddate('1970-01-01',t4*10000 + t3*1000 + t2*100 + t1*10 + t0) event_date 
    FROM
      (SELECT 0 t0 UNION SELECT 1 UNION SELECT 2 UNION SELECT 3 UNION SELECT 4 UNION SELECT 5 UNION SELECT 6 UNION SELECT 7 UNION SELECT 8 UNION SELECT 9) t0,
      (SELECT 0 t1 UNION SELECT 1 UNION SELECT 2 UNION SELECT 3 UNION SELECT 4 UNION SELECT 5 UNION SELECT 6 UNION SELECT 7 UNION SELECT 8 UNION SELECT 9) t1,
      (SELECT 0 t2 UNION SELECT 1 UNION SELECT 2 UNION SELECT 3 UNION SELECT 4 UNION SELECT 5 UNION SELECT 6 UNION SELECT 7 UNION SELECT 8 UNION SELECT 9) t2,
      (SELECT 0 t3 UNION SELECT 1 UNION SELECT 2 UNION SELECT 3 UNION SELECT 4 UNION SELECT 5 UNION SELECT 6 UNION SELECT 7 UNION SELECT 8 UNION SELECT 9) t3,
      (SELECT 0 t4 UNION SELECT 1 UNION SELECT 2 UNION SELECT 3 UNION SELECT 4 UNION SELECT 5 UNION SELECT 6 UNION SELECT 7 UNION SELECT 8 UNION SELECT 9) t4
  ) AS date_range
  JOIN 
    classes c
  JOIN 
    enrollment_changes AS e
  ON
    e.change_date <= date_range.event_date
  AND
    e.class_id = c.id
WHERE 
  date_range.event_date BETWEEN '2019-01-01' AND '2019-01-03'
;

当前结果：

event_date   name   NetEnrollment
------------------------------------
2019-01-01   math      6

想要的结果：

event_date   name      NetEnrollment
------------------------------------
2019-01-01   math      1
2019-01-01   english   0
2019-01-01   science   0
2019-01-02   math      1
2019-01-02   english   1
2019-01-02   science   0
2019-01-03   math      2
2019-01-03   english   1
2019-01-03   science   1

Answer 1

您可以使用以下解决方案：

SELECT date_range.event_date, c.name, IFNULL(SUM(e.change), 0) AS 'NetEnrollment'
FROM (
  SELECT adddate('1970-01-01',t4*10000 + t3*1000 + t2*100 + t1*10 + t0) event_date 
  FROM
    (SELECT 0 t0 UNION SELECT 1 UNION SELECT 2 UNION SELECT 3 UNION SELECT 4 UNION SELECT 5 UNION SELECT 6 UNION SELECT 7 UNION SELECT 8 UNION SELECT 9) t0,
    (SELECT 0 t1 UNION SELECT 1 UNION SELECT 2 UNION SELECT 3 UNION SELECT 4 UNION SELECT 5 UNION SELECT 6 UNION SELECT 7 UNION SELECT 8 UNION SELECT 9) t1,
    (SELECT 0 t2 UNION SELECT 1 UNION SELECT 2 UNION SELECT 3 UNION SELECT 4 UNION SELECT 5 UNION SELECT 6 UNION SELECT 7 UNION SELECT 8 UNION SELECT 9) t2,
    (SELECT 0 t3 UNION SELECT 1 UNION SELECT 2 UNION SELECT 3 UNION SELECT 4 UNION SELECT 5 UNION SELECT 6 UNION SELECT 7 UNION SELECT 8 UNION SELECT 9) t3,
    (SELECT 0 t4 UNION SELECT 1 UNION SELECT 2 UNION SELECT 3 UNION SELECT 4 UNION SELECT 5 UNION SELECT 6 UNION SELECT 7 UNION SELECT 8 UNION SELECT 9) t4
) AS date_range JOIN classes c 
  LEFT JOIN enrollment_changes AS e ON e.change_date <= date_range.event_date AND  c.id = e.class_id
WHERE date_range.event_date BETWEEN '2019-01-01' AND '2019-01-03'
GROUP BY date_range.event_date, c.name 
ORDER BY date_range.event_date, c.name

demo on dbfiddle.uk

日历上的每一天 table 与所有 class 一起加入（使用 JOIN）。注册更改加入（使用 LEFT JOIN）到当天和特定的 class。使用 IFNULL 您可以将 NULL 值替换为 0.

仅在当天使用 GROUP BY 会导致每天一行。要每天每 class 获得一行，您必须 GROUP BY 天和 class 姓名。

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