如何在SQL中找到没有聚合函数的最小值？_Sql_Postgresql_Group By_Greatest N Per Group_Metabase

如何在SQL中找到没有聚合函数的最小值？

sql postgresql

如何在SQL中找到没有聚合函数的最小值？,sql,postgresql,group-by,greatest-n-per-group,metabase,Sql,Postgresql,Group By,Greatest N Per Group,Metabase,我在Postgres上有一个示例数据集 updated_at activated_at name gender role school app_name device_type -------------+-----------------+----------+--------+-------+---------+---------+--------------- August 2 July 30 Ron

我在Postgres上有一个示例数据集

 updated_at     activated_at        name     gender   role    school  app_name    device_type
-------------+-----------------+----------+--------+-------+---------+---------+---------------
August 2         July 30            Ron        M       S        A         Y          android
August 1         July 30            Ron        M       S        A         Z          browser
July 30          July 30            Ron        M       S        A         Y          android
August 1         July 28            Ana        F       S        B         Y          android
August 1         July 28            Ana        F       S        B         Z          browser
July 28          July 28            Ana        F       S        B         Y          android

我想知道用户在激活后第一次显示（更新）的时间是什么时候，属于哪个应用程序

预期结果：

 updated_at     activated_at        name     gender   role    school  app_name    device_type
-------------+-----------------+----------+--------+-------+---------+---------+---------------
July 30          July 30            Ron        M       S        A         Y          android
July 28          July 28            Ana        F       S        B         Y          android

SELECT min(ut.updated_at), u.activated_at, u.full_name, u.gender, r.name, s.name, ut.app_name, ut.device_type
FROM "public"."user_tokens" ut JOIN
     "public"."users" u
     ON ut.user_id = u.id JOIN
     "public"."user_roles" ur
     ON ut.user_id = ur.user_id JOIN
     "public"."roles" r
     ON ur.role_id = r.id JOIN
     "public"."schools" s
      ON ur.school_id = s.id
WHERE (NOT (ut.app_name) like 'G')
Group by u.activated_at, u.full_name, u.gender, r.name, s.name, ut.app_name, ut.device_type
Order by u.activated_at desc

 updated_at     activated_at        name     gender   role    school  app_name    device_type
-------------+-----------------+----------+--------+-------+---------+---------+---------------
August 1         July 30            Ron        M       S        A         Z          browser
July 30          July 30            Ron        M       S        A         Y          android
August 1         July 28            Ana        F       S        B         Z          browser
July 28          July 28            Ana        F       S        B         Y          android

我尝试过以下SQL:

 updated_at     activated_at        name     gender   role    school  app_name    device_type
-------------+-----------------+----------+--------+-------+---------+---------+---------------
July 30          July 30            Ron        M       S        A         Y          android
July 28          July 28            Ana        F       S        B         Y          android

SELECT min(ut.updated_at), u.activated_at, u.full_name, u.gender, r.name, s.name, ut.app_name, ut.device_type
FROM "public"."user_tokens" ut JOIN
     "public"."users" u
     ON ut.user_id = u.id JOIN
     "public"."user_roles" ur
     ON ut.user_id = ur.user_id JOIN
     "public"."roles" r
     ON ur.role_id = r.id JOIN
     "public"."schools" s
      ON ur.school_id = s.id
WHERE (NOT (ut.app_name) like 'G')
Group by u.activated_at, u.full_name, u.gender, r.name, s.name, ut.app_name, ut.device_type
Order by u.activated_at desc

 updated_at     activated_at        name     gender   role    school  app_name    device_type
-------------+-----------------+----------+--------+-------+---------+---------+---------------
August 1         July 30            Ron        M       S        A         Z          browser
July 30          July 30            Ron        M       S        A         Y          android
August 1         July 28            Ana        F       S        B         Z          browser
July 28          July 28            Ana        F       S        B         Y          android

但结果如下：

 updated_at     activated_at        name     gender   role    school  app_name    device_type
-------------+-----------------+----------+--------+-------+---------+---------+---------------
July 30          July 30            Ron        M       S        A         Y          android
July 28          July 28            Ana        F       S        B         Y          android

SELECT min(ut.updated_at), u.activated_at, u.full_name, u.gender, r.name, s.name, ut.app_name, ut.device_type
FROM "public"."user_tokens" ut JOIN
     "public"."users" u
     ON ut.user_id = u.id JOIN
     "public"."user_roles" ur
     ON ut.user_id = ur.user_id JOIN
     "public"."roles" r
     ON ur.role_id = r.id JOIN
     "public"."schools" s
      ON ur.school_id = s.id
WHERE (NOT (ut.app_name) like 'G')
Group by u.activated_at, u.full_name, u.gender, r.name, s.name, ut.app_name, ut.device_type
Order by u.activated_at desc

 updated_at     activated_at        name     gender   role    school  app_name    device_type
-------------+-----------------+----------+--------+-------+---------+---------+---------------
August 1         July 30            Ron        M       S        A         Z          browser
July 30          July 30            Ron        M       S        A         Y          android
August 1         July 28            Ana        F       S        B         Z          browser
July 28          July 28            Ana        F       S        B         Y          android

我试图从group by子句中排除

app_name

和

device_type

，但它说

错误：“ut.app_name”列必须出现在group by子句中或用于聚合函数中

你知道怎么解决吗？如有任何意见，将不胜感激。谢谢。

我认为在Postgres上使用DISTINCT ON可能是最好的方法：

SELECT DISTINCT ON (u.full_name)
    ut.updated_at,
    u.activated_at,
    u.full_name,
    u.gender,
    r.name,
    s.name,
    ut.app_name,
    ut.device_type
FROM "public"."user_tokens" ut
INNER JOIN "public"."users" u ON ut.user_id = u.id
INNER JOIN "public"."user_roles" ur ON ut.user_id = ur.user_id
INNER JOIN "public"."roles" r ON ur.role_id = r.id
INNER JOIN "public"."schools" s ON ur.school_id = s.id
WHERE NOT ut.app_name LIKE 'G'
ORDER BY
    u.full_name,
    ut.updated_at;

上面将为每个全名用户返回一条记录，对应于在时间更新的较早的

记录。
我认为在
上进行区分可能是在Postgres上进行此操作的最佳方法：
SELECT DISTINCT ON (u.full_name)
    ut.updated_at,
    u.activated_at,
    u.full_name,
    u.gender,
    r.name,
    s.name,
    ut.app_name,
    ut.device_type
FROM "public"."user_tokens" ut
INNER JOIN "public"."users" u ON ut.user_id = u.id
INNER JOIN "public"."user_roles" ur ON ut.user_id = ur.user_id
INNER JOIN "public"."roles" r ON ur.role_id = r.id
INNER JOIN "public"."schools" s ON ur.school_id = s.id
WHERE NOT ut.app_name LIKE 'G'
ORDER BY
    u.full_name,
    ut.updated_at;

上面将为每个全名用户返回一条记录，对应于在

时间更新的较早的

记录。
您可以先按升序（从旧到新）排序，并使用“限制”来限制要显示的记录数量
select * from TABLE_NAME order by updated_at asc LIMIT 2;

您可以先按升序（从旧到新）排序，并使用“限制”来限制要显示的记录数量
select * from TABLE_NAME order by updated_at asc LIMIT 2;

您还可以使用PostgreSQL中的row\u number（）
执行此操作。您也可以使用PostgreSQL中的row\u number（）
执行此操作