Question

我正在使用以下声明并获得我需要的几乎所有内容。另外，我想获得所有row_total结果的SUM。

    SELECT c.id AS campaign_id, c.amount AS campaign_amount, c.warning_trigger,
COUNT(cc.id) as code_count, DATEDIFF(c.end_date, CURDATE()) as days_remain,
SUM(c.amount) as row_total
FROM campaigns c
JOIN campaign_codes cc
LEFT JOIN partners p on p.id = 4
WHERE c.partner_id = 4 AND cc.status = 0 AND c.id = cc.campaign_id
GROUP BY c.id
ORDER BY campaign_amount ASC

这是一个示例结果，最后一列是我想要添加到一起的列：

    1 10.00 30 992 24 9920.00
    3 25.00 30 995 290 24875.00
    16 50.00 30 988 335 49400.00
    17 100.00 30 1000 335 100000.00

Answer 1

获得运行总和：

SELECT  c.id AS campaign_id, c.amount AS campaign_amount, c.warning_trigger,
        COUNT(cc.id) as code_count, DATEDIFF(c.end_date, CURDATE()) as days_remain,
        SUM(c.amount) as row_total, @r := @r + SUM(c.amount) AS running_sum
FROM    (
        SELECT  @r := 0
        ) vars,
        campaign c, …

将总计（以及所有其他聚合）作为附加记录：

SELECT  *
FROM    (
        SELECT  c.id AS campaign_id, c.amount AS campaign_amount, c.warning_trigger,
                COUNT(cc.id) as code_count, DATEDIFF(c.end_date, CURDATE()) as days_remain,
                SUM(c.amount) as row_total
        FROM    campaign c, … 
        GROUP BY
                c.id WITH ROLLUP
        ) q
ORDER BY
        campaign_amount ASC

要在其他字段中获得总和：

        SELECT  c.id AS campaign_id, c.amount AS campaign_amount, c.warning_trigger,
                COUNT(cc.id) as code_count, DATEDIFF(c.end_date, CURDATE()) as days_remain,
                SUM(c.amount) as row_total,
                (
                SELECT  SUM(c.amount)
                FROM    campaign c, …
                -- the same but without the GROUP BY and ORDER BY clauses. 
                )
        FROM    campaign c, …
        GROUP BY
                c.id
        ORDER BY
                campaign_amount ASC

Answer 2

封装您的查询，如：

SELECT SUM([row_total]) FROM (
Your query
) Source

从下面的评论中，您可以尝试...

Your Query
UNION ALL
SELECT '', '', '', '', '', SUM([row_total]) FROM (
Your query
) Source

union all允许您在结果集中追加一行，这将为您提供“总行数”。您可以更改空字段（单引号）以匹配您的数据类型，或为它们指定代表总行数的特殊值。

Answer 3

SELECT TB.campaign_id, TB.campaign_amount, TB.warning_trigger, TB.code_count,  TB.days_remain, TB.row_total, SUM(TB.row_total) GlobalTotal FROM (SELECT c.id AS campaign_id, c.amount AS campaign_amount, c.warning_trigger,
COUNT(cc.id) as code_count, DATEDIFF(c.end_date, CURDATE()) as days_remain,
SUM(c.amount) as row_total
FROM campaigns c
JOIN campaign_codes cc
LEFT JOIN partners p on p.id = 4
WHERE c.partner_id = 4 AND cc.status = 0 AND c.id = cc.campaign_id
GROUP BY c.id
ORDER BY campaign_amount ASC) TB
GROUP BY TB.campaign_id, TB.campaign_amount, TB.warning_trigger, TB.code_count,  TB.days_remain, TB.row_total

MySQL - 如何获得COUNT字段的SUM？

3 个答案: