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，我正在做sum()一些值：
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，因此向count(vi_pdv_id_pdv)我返回一个值，该值计算所有这些重复的ID。我希望在vi_pdv_id_pdv不重复的情况下获得这些产品的价值。这里有一个小例子。通过此查询，我得到以下结果：
My Product | 32031 | 1098
在此的示例1098计算重复的值。我想得到：
My Product | 32031 | 534
该计数没有重复的值。我知道该group by语句不包含重复内容（如我已阅读的答案），但我无法实现与上面显示给您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
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