Mysql 如何除以两个计数语句？获取错误

我试图用礼物总数除以天数总数。Count语句是分开工作的，但是当我尝试划分这两个计数时，报告不会运行。有什么想法吗

((SELECT COUNT(*)
FROM ATTENDANCE
    LEFT OUTER JOIN ATTENDANCE_CODE
        ON ATTENDANCE_CODE.ID = ATTENDANCE.ATTENDANCE_CODEID
    LEFT OUTER JOIN PERIOD
        ON PERIOD.ID = ATTENDANCE.PERIODID
WHERE ATTENDANCE.STUDENTID = STUDENTS.ID
AND ATTENDANCE.ATT_DATE BETWEEN '%param1%' AND '%param2%'
AND PERIOD.PERIOD_NUMBER BETWEEN '1' AND '8'
AND ATTENDANCE.ATT_MODE_CODE = 'ATT_ModeMeeting'
AND ATTENDANCE.SCHOOLID = STUDENTS.SCHOOLID
AND ATTENDANCE_CODE.ATT_CODE = 'LZP'
)/
(SELECT COUNT(*)
FROM ATTENDANCE
    LEFT OUTER JOIN ATTENDANCE_CODE
        ON ATTENDANCE_CODE.ID = ATTENDANCE.ATTENDANCE_CODEID
    LEFT OUTER JOIN PERIOD
        ON PERIOD.ID = ATTENDANCE.PERIODID
WHERE ATTENDANCE.STUDENTID = STUDENTS.ID
AND ATTENDANCE.ATT_DATE BETWEEN '%param1%' AND '%param2%'
AND PERIOD.PERIOD_NUMBER BETWEEN '1' AND '8'
AND ATTENDANCE.ATT_MODE_CODE = 'ATT_ModeMeeting'
AND ATTENDANCE.SCHOOLID = STUDENTS.SCHOOLID
AND (ATTENDANCE_CODE.ATT_CODE = 'A' or ATTENDANCE_CODE.ATT_CODE = 'LZA' or ATTENDANCE_CODE.ATT_CODE = 
'LZP') 
))

---

您需要

选择该除法的结果。在分母中使用
NULLIF（）
也是一个好主意，以避免潜在的零除错误

SELECT
    (SELECT COUNT(*)
    FROM ATTENDANCE
        LEFT OUTER JOIN ATTENDANCE_CODE
            ON ATTENDANCE_CODE.ID = ATTENDANCE.ATTENDANCE_CODEID
        LEFT OUTER JOIN PERIOD
            ON PERIOD.ID = ATTENDANCE.PERIODID
    WHERE ATTENDANCE.STUDENTID = STUDENTS.ID
    AND ATTENDANCE.ATT_DATE BETWEEN '%param1%' AND '%param2%'
    AND PERIOD.PERIOD_NUMBER BETWEEN '1' AND '8'
    AND ATTENDANCE.ATT_MODE_CODE = 'ATT_ModeMeeting'
    AND ATTENDANCE.SCHOOLID = STUDENTS.SCHOOLID
    AND ATTENDANCE_CODE.ATT_CODE = 'LZP'
    )/
    NULLIF((SELECT COUNT(*)
    FROM ATTENDANCE
        LEFT OUTER JOIN ATTENDANCE_CODE
            ON ATTENDANCE_CODE.ID = ATTENDANCE.ATTENDANCE_CODEID
        LEFT OUTER JOIN PERIOD
            ON PERIOD.ID = ATTENDANCE.PERIODID
    WHERE ATTENDANCE.STUDENTID = STUDENTS.ID
    AND ATTENDANCE.ATT_DATE BETWEEN '%param1%' AND '%param2%'
    AND PERIOD.PERIOD_NUMBER BETWEEN '1' AND '8'
    AND ATTENDANCE.ATT_MODE_CODE = 'ATT_ModeMeeting'
    AND ATTENDANCE.SCHOOLID = STUDENTS.SCHOOLID
    AND (ATTENDANCE_CODE.ATT_CODE = 'A' or ATTENDANCE_CODE.ATT_CODE = 'LZA' or ATTENDANCE_CODE.ATT_CODE = 
    'LZP') 
    ), 0)

看看这两个子查询，它们非常相似，我想知道整个事情是否可以简化为：

SELECT AVG(ATTENDANCE_CODE.ATT_CODE = 'LZP')
FROM ATTENDANCE
LEFT OUTER JOIN ATTENDANCE_CODE
    ON ATTENDANCE_CODE.ID = ATTENDANCE.ATTENDANCE_CODEID
LEFT OUTER JOIN PERIOD
    ON PERIOD.ID = ATTENDANCE.PERIODID
WHERE 
    ATTENDANCE.STUDENTID = STUDENTS.ID
    AND ATTENDANCE.ATT_DATE BETWEEN '%param1%' AND '%param2%'
    AND PERIOD.PERIOD_NUMBER BETWEEN '1' AND '8'
    AND ATTENDANCE.ATT_MODE_CODE = 'ATT_ModeMeeting'
    AND ATTENDANCE.SCHOOLID = STUDENTS.SCHOOLID
    AND ATTENDANCE_CODE.ATT_CODE IN ('LZP', 'LZA', 'A')

不要选择两个
查询，而是使用一个查询，对不同的条件求和，然后将它们分开

SELECT SUM(ATTENDANCE_CODE.ATT_CODE = 'LZP') / SUM(ATTENDANCE_CODE.ATT_CODE IN ('A', 'LZA', 'LZP') AS attendance_ratio
FROM ATTENDANCE
INNER JOIN ATTENDANCE_CODE
    ON ATTENDANCE_CODE.ID = ATTENDANCE.ATTENDANCE_CODEID
INNER JOIN PERIOD
    ON PERIOD.ID = ATTENDANCE.PERIODID
WHERE ATTENDANCE.STUDENTID = STUDENTS.ID
AND ATTENDANCE.ATT_DATE BETWEEN '%param1%' AND '%param2%'
AND PERIOD.PERIOD_NUMBER BETWEEN '1' AND '8'
AND ATTENDANCE.ATT_MODE_CODE = 'ATT_ModeMeeting'
AND ATTENDANCE.SCHOOLID = STUDENTS.SCHOOLID

另外，看起来您不应该在这里使用
左外连接<代码>考勤\u代码ID
和
周期ID
似乎是外键，它们应该总是在父表中找到匹配项。
这不起作用，不确定是否是因为我还有其他查询。上面是学生信息。我添加了“（）”来解决这个问题，但仍然没有得出结果。谢谢你的努力！顶部似乎起了作用，但它给了我百分之一的折扣。因此，对于100%，我得到0.1，对于92，我得到9.247843002942（继续），不遵循代码中的无小数规则。有什么想法吗？非常感谢你！当我添加一个*100来解决这个问题时，它只适用于0.1，任何其他数字都不会更新