Hybris 关系型 FlexibleSearch 选择
Hybris relational FlexibleSearch selection
Hybris 为同一基本商店的用户创建多个购物车。如何 select 每个商店有超过一个购物车的用户?
稍后我会对它们进行排序并删除最旧的。
SELECT {u.code} as userCode,
{c.code} as cartCode,
{c.site} as cartSite
FROM {User as u join Cart as c on {u.pk} = {c.user}}
WHERE...
使用这个:
select {u.pk}, {b.pk}, count(distinct({c.pk}))
from {
user as u
join cart as c on {c.user} = {u.pk}
join basestore as b on {b.pk} = {c.store}
}
group by {u.pk}, {b.pk}
having count(distinct({c.pk})) > 1
如果您使用保存的购物车和报价功能,那么您可以考虑通过添加 WHERE {c.saveTime} IS NULL AND {c.quoteReference} IS NULL
不将其包含在结果中
select {u.pk}, {b.pk}, count(distinct({c.pk}))
from {
user as u
join cart as c on {c.user} = {u.pk}
join basestore as b on {b.pk} = {c.store}
}
WHERE {c.saveTime} IS NULL AND {c.quoteReference} IS NULL
GROUP BY {u.pk}, {b.pk}
having count(distinct({c.pk})) > 1
更新:
获取用户的唯一列表:
select {u.pk}
from {...
Hybris 为同一基本商店的用户创建多个购物车。如何 select 每个商店有超过一个购物车的用户?
稍后我会对它们进行排序并删除最旧的。
SELECT {u.code} as userCode,
{c.code} as cartCode,
{c.site} as cartSite
FROM {User as u join Cart as c on {u.pk} = {c.user}}
WHERE...
使用这个:
select {u.pk}, {b.pk}, count(distinct({c.pk}))
from {
user as u
join cart as c on {c.user} = {u.pk}
join basestore as b on {b.pk} = {c.store}
}
group by {u.pk}, {b.pk}
having count(distinct({c.pk})) > 1
如果您使用保存的购物车和报价功能,那么您可以考虑通过添加 WHERE {c.saveTime} IS NULL AND {c.quoteReference} IS NULL
select {u.pk}, {b.pk}, count(distinct({c.pk}))
from {
user as u
join cart as c on {c.user} = {u.pk}
join basestore as b on {b.pk} = {c.store}
}
WHERE {c.saveTime} IS NULL AND {c.quoteReference} IS NULL
GROUP BY {u.pk}, {b.pk}
having count(distinct({c.pk})) > 1
更新:
获取用户的唯一列表:
select {u.pk}
from {...