SQL查询是不是尽管使用IFNULL返回NULL值（MySQL的）

Question

我有一个测试台设置为这样：Table Rows setup

我的目标是尝试并获得当前部门以前设立该部门的计数。我的SQL是：

SELECT A.Department, IFNULL(COUNT(*), 0) 
FROM Departments A
INNER JOIN Departments B ON B.YearOfEstablishment < A.YearOfEstablishment
GROUP BY Department
ORDER BY COUNT(*);

但是我都试过LEFT JOIN和INNER JOIN，这是第一次发现从未返回，因为我可以假设它是无效的最后一个部门。尽管有IFNULL，该部门未显示。

我在做什么错在这里？

Answer 1

我认为这是你需要的查询：

SELECT A.Department, COUNT(B.Department)
FROM Departments A
LEFT JOIN Departments B ON B.YearOfEstablishment < A.YearOfEstablishment
GROUP BY A.Department
ORDER BY 2;

见this db fiddle demo。

| Department        | cnt |
| ----------------- | --- |
| Office Management | 0   |
| Business          | 1   |
| Sales Management  | 2   |
| ComputerScience   | 3   |
| Liberal Arts      | 4   |
| Farming           | 4   |
| Communications    | 6   |
| Digital Science   | 7   |

注：如由@fifonik评论，并不需要IFNULL因为COUNT已经返回0时没有记录可用。

Answer 2

在MySQL 8+，你可以更好地做到这一点使用rank()或row_number()：

SELECT d.Department,
       ROW_NUMBER() OVER (ORDER BY YearOfEstablishment) - 1 as seqnum
FROM Departments d
ORDER BY seqnum;

由于没有联系，这将是与您的查询。这可能是更好的事情可做：

SELECT d.Department,
       RANK() OVER (ORDER BY YearOfEstablishment) - 1 as seqnum
FROM Departments d
ORDER BY seqnum;

这应该是您正在寻找的计数。

Answer 3

你也可以做这样的事情：

SELECT
      A.Department
    , A.YearOfEstablishment
    , COUNT(*) - 1
FROM
    Departments A
    INNER JOIN Departments B ON (
           A.id = B.id
        OR B.YearOfEstablishment < A.YearOfEstablishment
    )
GROUP BY
    Department
ORDER BY
    COUNT(*);