Question

我的表结构：

consumer_id, signup_date, plan_id, subscription_date

它有相同consumer_id的多个subscription_dates。

我在结果中需要的是具有first（min）subscription_date的所有字段，其中plan_id不是特定值。

没有我试过的where子句：

select consumer_id, 
       signup_date, 
       plan_id, 
       min(subscription_date) 
from myTable 
Group by consumer_id

我收到错误声明

列＆＃39; signup_date＆＃39;，＆＃39; plan_id＆＃39;在选择列表中无效，因为它不包含在聚合函数或GROUP BY中子句。

但我不希望将结果分组为＆＃39; signup_date＆＃39;或者＆＃39; plan_id＆＃39;。

＆＃39; signup_date＆＃39;行中的值是多少？和＆＃39; plan_id＆＃39;在我获得每个consumer_id的最早的subscription_date之后，这就是我想要的值。如何在SQL中编写此条件？

Answer 1

你可以试试这个：

select consumer_id, signup_date, plan_id
     , min(subscription_date) over (partition by consumer_id)
from myTable

如果需要，您可以使用您的位置删除不需要的行。

Answer 2

如果没有其他行具有相同的consumer_id但存在更早的日期，请使用NOT EXISTS返回一行：

select consumer_id, 
       signup_date, 
       plan_id, 
       subscription_date
from myTable t1
where not exists (select 1 from myTable t2
                  where t2.consumer_id = t1.consumer_id
                    and t2.subscription_date < t1.subscription_date)

如果有两个来自相同的，最早的日期，则返回两行。

Answer 3

尝试此查询

select a.consumer_id, 
       a.signup_date, 
       a.plan_id,
       b.subcription_date
from myTable as a
join( 
select consumer_id, 
       min(subscription_date)  as subcription_date
from myTable
Group by consumer_id  ) b on a.subcription_date=b.subcription_date and a.consumer_id=b.consumer_id

Answer 4

select consumer_id, 
       signup_date, 
       plan_id, 
       min(subscription_date) 
from myTable 
Group by consumer_id, signup_date, plan_id

使用早期日期值为每个用户获取行

4 个答案: