【发布时间】：2015-07-18 12:48:45
【问题描述】：
以下查询为我提供了几乎正确的计数，除了它包含具有重复 p.promo_code 和 submitid 值的行。如何让这个查询使用 DISTINCT p.promo_code, submitid？
我尝试了一些我在 SO 上看到的用于单连接查询的解决方案，例如 --- 总和 * COUNT(DISTINCT p.promo_code, submitid)/count(*)
没有运气。 这是完整的查询：
SELECT
SUM(CASE WHEN ISNULL(p.promo_code)
AND h.account_number = c2.account_number THEN 1 ELSE 0 END) as 'no_match_acct_match',
SUM(CASE WHEN ISNULL(p.promo_code)
AND (h.account_number <> c2.account_number OR ISNULL(c2.account_number)) THEN 1 ELSE 0 END) as 'no_match',
SUM(CASE WHEN p.promo_code IS NOT NULL
AND c.full_name = CONCAT(h.firstname,' ',h.lastname)
AND p.start_date > @today THEN 1 ELSE 0 END) as 'future_name',
SUM(CASE WHEN p.promo_code IS NOT NULL
AND c.full_name = CONCAT(h.firstname,' ',h.lastname)
AND p.start_date <= @today
AND p.`status` = 'REDEEMED' THEN 1 ELSE 0 END) as 'redeemed_name',
SUM(CASE WHEN p.promo_code IS NOT NULL
AND c.full_name = CONCAT(h.firstname,' ',h.lastname)
AND p.start_date <= @today
AND p.`status` <> 'REDEEMED'
AND p.end_date < @today
AND p.end_date + 259200 >= @today THEN 1 ELSE 0 END) as 'buffered_name',
SUM(CASE WHEN p.promo_code IS NOT NULL
AND c.full_name = CONCAT(h.firstname,' ',h.lastname)
AND p.start_date <= @today
AND p.`status` <> 'REDEEMED'
AND p.end_date < @today
AND p.end_date + 259200 < @today THEN 1 ELSE 0 END) as 'expired_name',
SUM(CASE WHEN p.promo_code IS NOT NULL
AND c.full_name = CONCAT(h.firstname,' ',h.lastname)
AND p.start_date <= @today
AND p.`status` <> 'REDEEMED'
AND p.end_date >= @today THEN 1 ELSE 0 END) as 'current_name',
SUM(CASE WHEN p.promo_code IS NOT NULL
AND c.full_name <> CONCAT(h.firstname,' ',h.lastname)
AND p.start_date > @today
AND (r.resolve_status = 0 OR ISNULL(r.resolve_status)) THEN 1 ELSE 0 END) as 'future_noname',
SUM(CASE WHEN p.promo_code IS NOT NULL
AND c.full_name <> CONCAT(h.firstname,' ',h.lastname)
AND p.start_date <= @today
AND p.`status` = 'REDEEMED' THEN 1 ELSE 0 END) as 'redeemed_noname',
SUM(CASE WHEN p.promo_code IS NOT NULL
AND c.full_name <> CONCAT(h.firstname,' ',h.lastname)
AND p.start_date <= @today
AND p.`status` <> 'REDEEMED'
AND p.end_date < @today
AND p.end_date + 259200 >= @today THEN 1 ELSE 0 END) as 'buffered_noname',
SUM(CASE WHEN p.promo_code IS NOT NULL
AND c.full_name <> CONCAT(h.firstname,' ',h.lastname)
AND p.start_date <= @today
AND p.`status` <> 'REDEEMED'
AND p.end_date < @today
AND p.end_date + 259200 < @today THEN 1 ELSE 0 END) as 'expired_noname',
SUM(CASE WHEN p.promo_code IS NOT NULL
AND c.full_name <> CONCAT(h.firstname,' ',h.lastname)
AND p.start_date <= @today
AND p.`status` <> 'REDEEMED'
AND p.end_date >= @today THEN 1 ELSE 0 END) as 'current_noname'
FROM helpdesk h
FORCE INDEX (help_status)
LEFT JOIN promo_codes p
ON h.promo_code = p.promo_code
LEFT JOIN customer_promocode cp
ON p.promo_code = cp.promo_code
LEFT JOIN customers c
ON cp.customer_number = c.customer_number
LEFT JOIN customers c2
ON h.account_number = c2.account_number
LEFT OUTER JOIN redeem_queue r
ON p.promo_code = r.promo_code
WHERE
h.help_status = 0
AND h.in_view_by = ''
AND (p.start_date >= UNIX_TIMESTAMP(SUBDATE(CURDATE(), INTERVAL 1 YEAR))
OR p.start_date IS NULL)
AND (ISNULL(p.promo_code) OR ISNULL(r.promo_code))
它运行很快，解释计划看起来不错，但由于重复数据，目前有 2 个计数被取消。
我们将不胜感激。
【问题讨论】：
-
在没有数据的情况下要消化很多 SQL，但是您是否尝试过
GROUP BY p.promo_code？请注意，由于您的SUM中有CASE，我不能 100% 确定它是否是您要查找的内容 -
感谢@charmeleon，但使用 group by 会将数据分解为每个 promo_code 的记录。此查询正在获取每个 CASE 的 SUM。我知道在没有数据的情况下要查看和理解很多东西。如果我不能很快弄清楚，或者在这里找到解决方案，我会发布 DDL 以及每个表中的一小部分数据。