Question

我有一个SELECT语句，用于学校的学生列表，它目前使用以下代码运行，但我发现每次计数需要1-2秒，对此数据库的大多数其他简单查询不到一半一秒钟，所以我觉得我在这里效率很低。有一个更好的方法吗？ 1-2秒可能听起来不是很多，但在更详细的声明中我添加到下面还要计算出现/缺席的主题等，所以当我在声明中包含所有内容时，它将花费近30秒来运行。

        SELECT studentID AS sID, stu.name AS Name,
        (
        SELECT COUNT( * )
        FROM attendance
        WHERE ((sID = 1)
        AND classdate < CURDATE( ))
        ) AS present, 
        (
        SELECT COUNT( * )
        FROM attendance
        WHERE ((sID != 1)
        AND classdate < CURDATE( ))
        ) AS absent
        FROM attendance
        INNER JOIN students AS stu ON attendance.studentID = stu.id
        WHERE awayid != -1
        GROUP BY sID
        ORDER BY present ASC

        ---------------------------------------------
        |   sID |   Name    |   present |   absent  |
        |   1   |   John    |   28      |   2       |
        |   2   |   Richard |   26      |   4       |
        |   3   |   Harry   |   22      |   8       |
        |   4   |   Sarah   |   22      |   8       |
        |   5   |   Tanya   |   20      |   10      |
        |   6   |   Nathan  |   20      |   10      |
        |   7   |   Kate    |   20      |   10      |
        ---------------------------------------------

Answer 1

做一些假设，试试这个：

SELECT
    studentID AS sID,
    stu.name AS Name,
    sum(a.sID = 1) as present,
    sum(a.sID != 1) as absent
FROM students AS stu
JOIN attendance a ON a.studentID = stu.id
    AND classdate < CURDATE( )
WHERE awayid != -1
GROUP BY 1, 2
ORDER BY present ASC

请注意，在mysql中，true为1，false为0，因此SUM(condition)计算它的真实次数，从而避免繁琐地使用CASE其他数据库要求。

Answer 2

您可以使用条件聚合：

select studentID as sID, stu.name,
       sum((sID = 1) AND classdate < CURDATE( )) as present,
       sum((sID != 1) AND classdate < CURDATE( )) as absent
FROM attendance INNER JOIN
     students AS stu
     ON attendance.studentID = stu.id
WHERE awayid != -1
GROUP BY sID
ORDER BY present ASC;

但让我感到困惑的是在子查询中使用sID。它不会引用外studentId中的select。我怀疑应该在那里使用另一个变量。（并且，如果sid中有attendance，您为什么要在外部查询中调用studentID sID？）

更有效的方法来计算mysql中的数据组？

2 个答案: