SQL Server：在 Group by 中显示 0 并具有不同的值

Question

我有一张每天有很多条目的表格。每个条目都有一个“级别”。现在我要做的是，每天选择一个级别存在多少条记录。级别可以不同，所以首先我需要选择不同的所有级别，并每天显示每个级别有多少记录。如果一天没有按级别记录，则该值应为0。

为了说明我的需求： 我的表有以下内容

¦Date ¦ Level ¦ RecId ¦ ------------------------------ ¦2014-06-18 ¦ 1 ¦ 1 ¦2014-06-18 ¦ 1 ¦ 2 ¦2014-06-18 ¦ 2 ¦ 3 ¦2014-11-06 ¦ 1 ¦ 4 ¦2014-11-11 ¦ 2 ¦ 5 ¦2014-11-13 ¦ 3 ¦ 6

我想要的是：

¦ 日期 ¦ 级别 ¦ CountOfRecords ¦ -------------------------------------- ¦2014-06-18 ¦ 1 ¦ 2 ¦2014-06-18 ¦ 2 ¦ 1 ¦2014-08-18 ¦ 3 ¦ 0 ¦2014-11-06 ¦ 1 ¦ 1 ¦2014-11-06 ¦ 2 ¦ 0 ¦2014-11-06 ¦ 3 ¦ 0 ¦2014-11-11 ¦ 1 ¦ 0 ¦2014-11-11 ¦ 2 ¦ 1 ¦2014-11-11 ¦ 3 ¦ 0 ¦2014-11-13 ¦ 1 ¦ 0 ¦2014-11-13 ¦ 2 ¦ 0 ¦2014-11-13 ¦ 3 ¦ 1

我需要它的原因是在折线图中显示这些值，并且每个级别都是一条线。

感谢您的帮助，而我仍在继续使用 google...

干杯

Answer 1

试试这个：

    SELECT T.[Date], 
           T.[Level], 
           COALESCE(COUNT(D.[Level]),0) AS CountOfRecords
    FROM (
          SELECT DISTINCT B.Date, A.[Level] 
          FROM tablename AS A 
                INNER JOIN (SELECT DISTINCT [Date] 
                            FROM tablename) AS B ON 1=1
          ) 
          AS T LEFT JOIN tablename AS D ON T.[Level] = D.[Level] 
                                           AND T.[Date] = D.[Date]
    GROUP BY T.[Date], T.[Level]
    ORDER BY T.[Date], T.[Level]

test is here

Answer 2

在实际表上产生 cross join 和 dates 和 levels，然后是 left join：

;with cte1 as(select distinct level from tablename),
      cte2 as(select distinct date from tablename)
select c1.level, 
       c2.date,
       count(distinct t.RecId) as CountOfRecords 
from cte1 c1
cross join cte2 c2
left join tablename t on t.date = c2.date and t.level = c1.level
group by c1.level, c2.date

编辑：

DECLARE @t TABLE
    (
      Date DATE ,
      Level INT ,
      RecId INT
    )

INSERT  INTO @t
VALUES  ( '2014-06-18', 1, 1 ),
        ( '2014-06-18', 1, 2 ),
        ( '2014-06-18', 2, 3 ),
        ( '2014-11-06', 1, 4 ),
        ( '2014-11-11', 2, 5 ),
        ( '2014-11-13', 3, 6 )

;with cte1 as(select distinct level from @t),
      cte2 as(select distinct date from @t)
select c1.level, 
       c2.date,
       count(distinct t.RecId) as CountOfRecords 
from cte1 c1
cross join cte2 c2
left join @t t on t.date = c2.date and t.level = c1.level
group by c1.level, c2.date

输出：

level   date    CountOfRecords
1       2014-06-18  2
2       2014-06-18  1
3       2014-06-18  0
1       2014-11-06  1
2       2014-11-06  0
3       2014-11-06  0
1       2014-11-11  0
2       2014-11-11  1
3       2014-11-11  0
1       2014-11-13  0
2       2014-11-13  0
3       2014-11-13  1

Answer 3

试试这个：

SELECT DISTINCT 
        [Date]
      , [level_ordinal] AS [level]
      , (SELECT COUNT([RecId]) 
            FROM [entries] AS E2 
            WHERE E2.[Level] = L.[level_ordinal]
                AND E2.[Date] = E1.[Date]) AS CountOfRecords
  FROM [entries] AS E1
  CROSS JOIN [entries_levels] AS L

与：

CREATE TABLE entries_levels
(
    [level_ordinal] [int] NULL
)

和

INSERT INTO entries_levels VALUES (1), (2), (3)

Answer 4

试试这个：

创建样本表：

CREATE TABLE NewTable
(
Date DATETIME,
Level INT,
RecID INT
)

插入样本数据：

INSERT INTO NewTable
VALUES
('2014-06-18', '1', '1'),
('2014-06-18', '1', '2'),
('2014-08-18', '2', '3'),
('2014-11-06', '1', '4'),
('2014-11-11', '2', '5'),
('2014-11-13', '3', '6')

SQL 查询：

SELECT
    Date,
    Level,
    COUNT(RecID) AS [CountOfRecords]
FROM NewTable
GROUP BY Date, Level

输出：

======================
|Date      |Lvl|Count|
|2014-06-18|1  |2    |
|2014-11-06|1  |1    |
|2014-08-18|2  |1    |
|2014-11-11|2  |1    |
|2014-11-13|3  |1    |
======================

PS：我认为您需要重新读取您的示例数据与您的预期输出，因为您已经迭代了日期 2014-06-18 的级别 1-3，但在一行中您有 2014-08-18 并继续作为3级所以计数不正确？