Question

我有一张类似波纹管的桌子。我想获取每天的访问用户数和角色。

表格：

CREATE TABLE #TempUser
(
  [Id] int NULL,
  [Username] nvarchar(50) NULL
)

CREATE TABLE #TempRole
(
  [Id] int NULL,
  [Title] nvarchar(50) NULL
)

CREATE TABLE #TempUserRole
(
  [UserId] int NULL,
  [RoleId] int NULL,
  [LastDate] date NULL
)

数据：

INSERT INTO #TempUser([Id],[UserName])
VALUES
(1,'Marvin'),
(2,'Tom'),
(3,'Charli')

INSERT INTO #TempRole([Id],[Title])
VALUES
(1,'Admin'),
(2,'Manager'),
(3,'Guest')

INSERT INTO #TempUserRole([UserId],[RoleId],[LastDate])
VALUES
(1,1,'2019-02-03'),
(1,2,'2019-02-06'),
(1,2,'2019-02-03'),
(3,3,'2019-02-03'),
(3,1,'2019-02-02'),
(3,1,'2019-02-03'),
(2,1,'2019-02-05'),
(2,1,'2019-02-05'),
(3,1,'2019-02-03'),
(1,1,'2019-02-03'),
(2,1,'2019-02-06'),
(2,1,'2019-02-03'),
(1,1,'2019-02-03'),
(2,1,'2019-02-02'),
(1,1,'2019-02-03'),
(2,1,'2019-02-05'),
(1,1,'2019-02-05'),
(2,1,'2019-02-03')

如何获取每天具有相同“用户”，“角色”，“日期”，“ CounterOfDay”的用户列表的访问用户数和角色？我需要的输出应如下所示：

UserName  Role    Date          CounterOfDay

Charli    Admin   2019-02-02    2

Tom       Admin   2019-02-02    2
...

Answer 1

这是您的答案，但是请检查它是否可能存在任何语法错误，因为我不太经常为SQL-Server写查询。

WITH counters AS (
   SELECT UserId, RoleId, LastDate, COUNT(*) AS CounterOfDay 
   FROM TempUserRole 
   GROUP BY UserId, RoleId, LastDate
)

SELECT 
   user.UserName
   , role.Title
   , counters.LastDate
   , counters.CounterOfDay
FROM counters
JOIN TempRole AS role ON counters.RoleId = TempRole.Id
JOIN TempUser AS user ON counters.UserId = TempUser.Id

我认为阅读起来非常不错，并且应该比第一次进行连接时表现更好。

Answer 2

您可以尝试以下操作。这是demo

with cte as
(
  select
    userName,
    title,
    lastDate,
    count(*) as total
  from tempUserRole tur
  join tempUser tu
  on tur.userId = tu.id

  join tempRole tr
  on tur.roleId = tr.id

  group by
    userName,
    title,
    lastDate
),
rnk as  
(
  select 
      *,
      dense_rank() over (order by lastDate, title, total) as rnk
  from cte
),
cnt as
( 
  select
    *,
    count(*) over (partition by rnk) as ttl
  from rnk
)


select
    userName,
    title,
    lastDate,
    total
from cnt
where ttl > 1

如何获得一些物品，然后删除重复的物品？

2 个答案: