MySQL SELECT MAX of SUM() 值答案

【问题标题】：MySQL SELECT MAX of SUM() valuesMySQL SELECT MAX of SUM() 值
【发布时间】：2012-11-26 01:29:45
【问题描述】：

我有这个 MySQL 查询：

SELECT 
    SUM(scpe.scpe_estemated_days) AS total_days,
    scp.cpl_startdate
FROM
    studentcourseplanelements scpe
        INNER JOIN
    studentcourseplan scp ON scp.cpl_id = scpe.scpe_cpl_id
        INNER JOIN
    (SELECT 
        sd1.student_id, sd1.student_startdate
    FROM
        studentdates sd1
    WHERE
        sd1.student_id = '360'
    LIMIT 1) sd ON sd.student_id = scp.student_id
GROUP BY scp.cpl_id

这个输出：

+------------+---------------+
| total_days | cpl_startdate |
+------------+---------------+
|          5 | 2012-11-01    |
|        129 | 2012-11-02    |
+------------+---------------+

我只想选择total_days 最高的行，在我的示例中为 129。

我该怎么做？

【问题讨论】：

标签： mysql sql sum max

【解决方案1】：

这将显示具有最高 total_days 的重复记录，例如

+------------+---------------+
| total_days | cpl_startdate |
+------------+---------------+
|          5 | 2012-11-01    |
|        129 | 2012-11-02    |  <= shown
|        129 | 2012-11-03    |  <= shown
+------------+---------------+

查询

SELECT  SUM(scpe.scpe_estemated_days) AS total_days,
        scp.cpl_startdate
FROM    studentcourseplanelements scpe INNER JOIN studentcourseplan scp
            ON scp.cpl_id = scpe.scpe_cpl_id
        INNER JOIN
        (SELECT  sd1.student_id, sd1.student_startdate
         FROM    studentdates sd1
         WHERE   sd1.student_id = '360'
         LIMIT 1) sd ON sd.student_id = scp.student_id
GROUP BY scp.cpl_id
HAVING SUM(scpe.scpe_estemated_days) = 
    (
        SELECT MAX(total_days)
        FROM
        (
            SELECT  SUM(scpe.scpe_estemated_days) AS total_days,
            FROM    studentcourseplanelements scpe INNER JOIN studentcourseplan scp 
                         ON scp.cpl_id = scpe.scpe_cpl_id
                    INNER JOIN
                    (SELECT  sd1.student_id, sd1.student_startdate
                     FROM    studentdates sd1
                     WHERE   sd1.student_id = '360'
                     LIMIT 1) sd ON sd.student_id = scp.student_id
            GROUP BY scp.cpl_id
        ) s
    )

【讨论】：

【解决方案2】：

尝试添加ORDER BY total_days DESC LIMIT 0,1

这将按大多数排序，并且只返回最高答案。

【讨论】：

谢谢！是的，您和斯蒂芬都为我提供了相同的答案，既然您确实是第一个，我会将其标记为答案。但是我必须再等 9 分钟才能这样做:)
是的，您是第一个，但我有一个问题，如果您有两条具有相同 total_days 的记录怎么办？例如5, 2012-01-01, 10, 2012-01-02,10, 2012-01-03 ?

【解决方案3】：

基本上，您希望按 SUM 列降序对其进行排序，然后获取列出的第一条记录。试试这个：

SELECT 
    SUM(scpe.scpe_estemated_days) AS total_days,
    scp.cpl_startdate
FROM
    studentcourseplanelements scpe
        INNER JOIN
    studentcourseplan scp ON scp.cpl_id = scpe.scpe_cpl_id
        INNER JOIN
    (SELECT 
        sd1.student_id, sd1.student_startdate
    FROM
        studentdates sd1
    WHERE
        sd1.student_id = '360'
    LIMIT 1) sd ON sd.student_id = scp.student_id
ORDER BY SUM(scpe.scpe_estemated_days) DESC
GROUP BY scp.cpl_id
LIMIT 1

【讨论】：

谢谢！但是 GROUP BY 出现在 ORDER BY 之前 :)
如果您有两条总天数相同的记录怎么办？例如5, 2012-01-01, 10, 2012-01-02,10, 2012-01-03 ?