我进行了以下查询:
select ba_brand, ceil(sum((sk_front + sk_depth + sk_warehouse + (sk_exhibition IS NOT NULL))) / count(distinct(vi_pdv_id_pdv))) as 'inventory',
count(vi_pdv_id_pdv)
from sf_visit_stock
join sf_visit on sk_vi_id_visit = id_visit
join sf_pdv on vi_pdv_id_pdv = id_pdv
join sf_format on pdv_fo_id_format = id_format
join sf_group on fo_gr_id_group = id_group
join sf_channel on gr_ch_id_channel = id_channel
join sf_product on sk_pd_id_product = id_product
join sf_family on pd_fa_id_family = id_family
join sf_brand on fa_ba_id_brand = id_brand
where (CASE WHEN from_unixtime(vi_scheduled_start,'%Y-%m-%d') between '2014-10-01' and '2014-10-31' AND vi_vs_id_visit_status in (1,2,3,4,6) THEN 4 END)
and vi_pr_id_proyect = 5 and ba_rival = 0 and id_brand in (72,75,76,77,78,79,80)
and sf_channel.ch_channel not in ("CHAINS DIRECTS", "CHAINS INDIRECTS", "INDEPENDENTS", "SUPERMARKET")
group by ba_brand
换句话说,我正在执行某些值的sum()
:
sum((sk_front + sk_depth + sk_warehouse + (sk_exhibition IS NOT NULL))
从许多行中:
count(vi_pdv_id_pdv)
但是这些行中的ID是重复的,因此count(vi_pdv_id_pdv)
给我返回一个值,该值计算所有这些重复的ID。我想获得这些vi_pdv_id_pdv
的值而不会重复。这里有一个小例子。通过此查询,我得到以下结果:
My Product | 32031 | 1098
在此示例中1098计算重复的值。我想得到:
My Product | 32031 | 534
该计数没有重复的值。我知道group by
语句不重复,就像我读过的this answer一样,但是我无法实现与上面显示的查询中类似的内容。
您可以使用COUNT(DISTINCT expr,[expr ...])
http://dev.mysql.com/doc/refman/5.6/en/group-by-functions.html#function_count-distinct
重复项来自哪里?
我假设它们是由您的一个连接引起的。您可以在此联接中添加其他子句以摆脱它们。示例:
SELECT *
FROM sf_visit_stock AS vs
JOIN sf_visit AS v ON vs.sk_vi_id_visit = v.id_visit AND v.somevalue = 1
也有可能在WHERE子句中摆脱它们。