COUNT(DISTINCT) 的百分位数与相关 WHERE 仅适用于视图(或没有 DISTINCT)

EDIT : There was a calculation error in my earlier code. Updated now.

SELECT COUNT(id) INTO @no_of_attempts from tb2;

-- DROP TABLE IF EXISTS S1Percentiles;
-- DROP TABLE IF EXISTS S2Percentiles;
-- DROP TABLE IF EXISTS S3Percentiles;

CREATE TEMPORARY TABLE S1Percentiles (
    s1 FLOAT NOT NULL,
    percentile FLOAT NOT NULL DEFAULT 0.00
);

CREATE TEMPORARY TABLE S2Percentiles (
    s2 FLOAT NOT NULL,
    percentile FLOAT NOT NULL DEFAULT 0.00
);

CREATE TEMPORARY TABLE S3Percentiles (
    s3 FLOAT NOT NULL,
    percentile FLOAT NOT NULL DEFAULT 0.00
);



INSERT INTO S1Percentiles (s1, percentile)
    SELECT A.s1, ((COUNT(B.s1)/@no_of_attempts)*100)
    FROM (SELECT DISTINCT s1 from tb2) A
    INNER JOIN tb2 B
    ON B.s1 <= A.s1
    GROUP BY A.s1;

INSERT INTO S2Percentiles (s2, percentile)
    SELECT A.s2, ((COUNT(B.s2)/@no_of_attempts)*100)
    FROM (SELECT DISTINCT s2 from tb2) A
    INNER JOIN tb2 B
    ON B.s2 <= A.s2
    GROUP BY A.s2;

INSERT INTO S3Percentiles (s3, percentile)
    SELECT A.s3, ((COUNT(B.s3)/@no_of_attempts)*100)
    FROM (SELECT DISTINCT s3 from tb2) A
    INNER JOIN tb2 B
    ON B.s3 <= A.s3
    GROUP BY A.s3;

-- select * from S1Percentiles;
-- select * from S2Percentiles;
-- select * from S3Percentiles;

UPDATE tb1 A
    INNER JOIN
    (
    SELECT B.tb1_id AS id, (C.percentile + D.percentile + E.percentile) AS sum FROM tb2 B
        INNER JOIN S1Percentiles C
        ON B.s1 = C.s1
        INNER JOIN S2Percentiles D
        ON B.s2 = D.s2
        INNER JOIN S3Percentiles E
        ON B.s3 = E.s3
    ) F
    ON A.id = F.id

    SET A.sum = F.sum;

-- SELECT * FROM tb1;

DROP TABLE S1Percentiles;
DROP TABLE S2Percentiles;
DROP TABLE S3Percentiles;

这样做是为了记录每个分数组的百分位数，然后最后只用必要的数据更新 tb1 列，而不是重新计算每个学生行的百分位数.