SQL：从列返回最频繁的值

我有一个名为customerOrders的下表

ID    user    order
1      1        2
2      1        3
3      1        1
4      2        1
5      1        5
6      2        4
7      3        1
8      6        2
9      2        2
10     2        3 

我想返回到订单最多的用户。目前，我有以下疑问：

SELECT user, COUNT(user) AS UsersWithMostOrders
FROM customerOrders
GROUP BY user
ORDER BY UsersWithMostOrders DESC;

这将返回按总订单分组的所有值，如

user     UsersWithMostOrders
1               4
2               4
3               1
6               1

我只想返回订单最多的用户。在我的例子中，这将是用户1和2，因为它们都有4个订单。如果我使用TOP 1或LIMIT，它将只返回第一个用户。如果我使用TOP 2，它将只在这种情况下工作，当前两个用户的订单数不同时，它将返回无效数据

要求的结果

您可以使用：

看。 结果:

您可以使用：

看。 结果:

选择1 应该适用于大多数版本的SQL

select * 
from (
  select    *, 
          rank() over(order by numOrders desc) as rrank
  from (
    select `user`, count(*) as numOrders
    from customerOrders
    group by `user`
  ) summed 
) ranked 
where rrank = 1

玩转代码

选择2 如果您的SQL版本允许使用窗口函数，那么这里有一个更具可读性的解决方案，它可以做同样的事情

with summed as (
    select `user`, count(*) as numOrders
    from customerOrders
    group by `user` 
),
ranked as (
    select *, 
           rank() over(order by numOrders desc) as rrank
    from summed 
)
select * 
from ranked 
where rrank = 1

使用代码玩转选项1 应该适用于大多数版本的SQL

select * 
from (
  select    *, 
          rank() over(order by numOrders desc) as rrank
  from (
    select `user`, count(*) as numOrders
    from customerOrders
    group by `user`
  ) summed 
) ranked 
where rrank = 1

玩转代码

选择2 如果您的SQL版本允许使用窗口函数，那么这里有一个更具可读性的解决方案，它可以做同样的事情

with summed as (
    select `user`, count(*) as numOrders
    from customerOrders
    group by `user` 
),
ranked as (
    select *, 
           rank() over(order by numOrders desc) as rrank
    from summed 
)
select * 
from ranked 
where rrank = 1

---

玩转代码

您可以使用CTE来达到以下要求：

;WITH CTE AS(
SELECT [user], COUNT(user) AS UsersWithMostOrders
FROM @T
GROUP BY [user])

SELECT M.* from CTE M
INNER JOIN ( SELECT
MAX(UsersWithMostOrders) AS MaximumOrders FROM CTE) S ON 
M.UsersWithMostOrders=S.MaximumOrders

---

您可以使用CTE来达到以下要求：

;WITH CTE AS(
SELECT [user], COUNT(user) AS UsersWithMostOrders
FROM @T
GROUP BY [user])

SELECT M.* from CTE M
INNER JOIN ( SELECT
MAX(UsersWithMostOrders) AS MaximumOrders FROM CTE) S ON 
M.UsersWithMostOrders=S.MaximumOrders

---

下面是Oracle查询的帮助：

WITH test_table AS
(
SELECT user, COUNT(order) AS total_order , DENSE_RANK() OVER (ORDER BY 
total_order desc) AS rank_orders  FROM customerOrders
GROUP BY user 
)
select * from test_table where rank_orders = 1

---

下面是Oracle查询的帮助：

WITH test_table AS
(
SELECT user, COUNT(order) AS total_order , DENSE_RANK() OVER (ORDER BY 
total_order desc) AS rank_orders  FROM customerOrders
GROUP BY user 
)
select * from test_table where rank_orders = 1

---

您使用的是哪个版本的SQL？MySQL、PostgreSQL、MSSQL等，并提供SQL版本将是有益的。您使用的是哪个版本的SQL？MySQL、PostgreSQL、MSSQL等，并提供SQL版本将是有益的