带有很多LEFT JOINS的SQL查询

Question

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        |

Answer 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,

Answer 2

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;