我正在尝试整理一个查询，但结果不是我所期望的，
select pii.id, pii.name, pii.sku, group_concat(pic.id) category_ids, group_concat(pimg.id) as modifier_group_ids
from items as pii
left join category_item as pici on pii.id = pici.item_id
left join categories as pic on pici.category_id = pic.id
left join item_modifier_group as piimg on pii.id = piimg.item_id
left join modifier_groups as pimg on piimg.modifier_group_id = pimg.id
where pii.account_id = 728
group by pii.id;
结果是这样，第4列（category_ids）被分组为重复值..
id |name |sku |category_ids |modifier_group_ids |
-------|----------------------------|----|------------------------|-------------------|
336937 |Chicken Pullets | |12981 | |
336938 |Hats | |12965 | |
336939 |Jack Fruit Taco Salad | |12975,12975,12975 |57,58,60 |
336940 |Vegatarian Taco Salad | |12975,12975,12975 |58,60,57 |
336941 |Pulled Pork Taco Salad | |12975,12975,12975 |58,57,60 |
336942 |Ground Beef Taco Salad | |12975,12975,12975 |60,57,58 |
336943 |Steak Taco Salad | |12975,12975,12975 |57,58,60 |
336944 |Chicken Taco Salad | |12975,12975,12975 |60,58,57 |
336945 |Jack Fruit Quesadilla | |12976,12976,12976,12976 |58,60,57,56 |
答案 0 :(得分：1)
您的查询看起来很好，而且大部分都是JOIN。请尝试使用distinct中的group_concat()，如下所示。
group_concat(distinct pic.id) category_ids,
答案 1 :(得分：0)
如果您不希望group_concat()连接重复值，则只需在函数中使用distinct关键字：
select pii.id, pii.name, pii.sku, group_concat(DISTINCT pic.id) category_ids, group_concat(pimg.id) as modifier_group_ids
from items as pii
left join category_item as pici on pii.id = pici.item_id
left join categories as pic on pici.category_id = pic.id
left join item_modifier_group as piimg on pii.id = piimg.item_id
left join modifier_groups as pimg on piimg.modifier_group_id = pimg.id
where pii.account_id = 728
group by pii.id;