使用 SQL 根据小时数求和小时值、计数和显示

Question

我有两张表，分别是教师和活动。

CREATE TABLE teacher (
TeacherId INT, BranchId VARCHAR(5));
INSERT INTO teacher VALUES
("1121","A"),
("1132","A"),
("1141","A"),
("2120","B"), 
("2122","B");
                               
CREATE TABLE activities (
ID INT, TeacherID INT,    Hours   INT);

INSERT INTO activities VALUES
(1,1121,2),
(2,1121,1),
(3,1132,1),
(4,1141,NULL),
(5,2120,NULL),
(6,2122,NULL);

NULL 表示没有活动，将在输出表上转换为 0。我想生成一个查询来计算总小时数并根据教师小时数计算有多少活动，如下表：

   +-----------+------------+------------+
   |   Hours   |     A      |     B      |
   +-----------+------------+------------+
   |     0     |     1      |     2      |
   |     1     |     1      |     0      |
   |     2     |     0      |     0      |
   |     3     |     1      |     0      |
   +-----------+------------+------------+

已编辑：对不起，我不知道如何准确地阐述，但这是我从其他成员https://www.db-fiddle.com/f/mmtuZquKyUqdhPvTFN9qaF/1收到的小提琴

编辑：最后，修改需要，将小时数相加并根据分支 ID 和教师 ID 计算小时数作为输出。

此处的预期输出（红色文本）：https://drive.google.com/file/d/1wyZ_aX5hz_7I1Ncf5sXLpstYk6FT8PMg/view?usp=sharing

Answer 1

我们可以通过使用加入聚合子查询的小时日历表来处理这个问题：

SELECT
    t1.Hours,
    SUM(CASE WHEN t2.BranchId = 'A' THEN t2.cnt ELSE 0 END) AS A,
    SUM(CASE WHEN t2.BranchId = 'B' THEN t2.cnt ELSE 0 END) AS B
FROM (SELECT 0 AS Hours UNION ALL SELECT 1 UNION ALL SELECT 2 UNION ALL SELECT 3) t1
LEFT JOIN
(
    SELECT t.BranchId, COALESCE(a.Hours, 0) AS Hours, COUNT(*) AS cnt
    FROM Teacher t
    LEFT JOIN Activities a ON a.TeacherId = t.TeacherId
    GROUP BY t.BranchId, COALESCE(a.Hours, 0)
) t2
    ON t1.Hours = t2.Hours
GROUP BY
    t1.Hours
ORDER BY
    t1.Hours

Demo

Answer 2

这基本上是一个 JOIN 和聚合。 . .但你需要从你想要的所有时间开始：

SELECT h.Hours,
       COALESCE(SUM(t.BranchId = 'A'), 0) AS A,
       COALESCE(SUM(t.BranchId = 'B'), 0) AS B
FROM (SELECT 0 AS Hours UNION ALL SELECT 1 UNION ALL SELECT 2 UNION ALL SELECT 3
     ) h LEFT JOIN
     activities a
     ON h.hours = COALESCE(a.hours, 0) LEFT JOIN
     teacher t
     ON t.TeacherId = a.TeacherId
GROUP BY h.Hours
ORDER BY h.Hours;

Here 是一个 dbfiddle。