SQL查询获取每月报告的计数项目？

Question

当我输入month和year时，我需要每月从表项的每天（固定day1-day31作为列）的表中进行报告的sql查询。

这是我的桌子（项目）

ID | NAME | 日期
-------------------------------------------------- --
1 | 项目A | 2015-2-25 13:37:49
2 | 项目A | 2015-2-25 14:37:49
3 | 项目A | 2015-2-26 13:30:55
4 | 项目B | 2015-2-26 15:37:49
5 | 项目B | 2015-2-26 17:57:49
6 | 项目C | 2015-2-27 13:00:33

---

（输入月份= 02和年份= 2015）
我需要用以下视图实现：

NAME | 1 | 2 | 3 |…| 25 | 26 | 27 | 28 | 29 | 30 | 31 |总计
-------------------------------------------------- ----
项目A | 0 | 0 | 0 |…| 2 | 1 | 0 | 0 | 0 | 0 | 0 | 3
项目B | 0 | 0 | 0 |…| 0 | 2 | 0 | 0 | 0 | 0 | 0 | 2
项目C | 0 | 0 | 0 |…| 0 | 0 | 1 | 0 | 0 | 0 | 0 | 1个

---

任何想法将不胜感激。

提前致谢。

抱歉，这是我的第一篇文章。

Answer 1

这将为您做到。 首次测试数据：

CREATE TABLE data ([ID] int, [Name] varchar(30), [Date] datetime)

INSERT INTO data ([ID], [Name], [Date])
SELECT 1,'ITEM A','2015-2-25 13:37:49'
UNION ALL SELECT 2,'ITEM A','2015-2-25 14:37:49'
UNION ALL SELECT 3,'ITEM A','2015-2-26 13:30:55'
UNION ALL SELECT 4,'ITEM B','2015-2-26 15:37:49'
UNION ALL SELECT 5,'ITEM B','2015-2-26 17:57:49'
UNION ALL SELECT 6,'ITEM C','2015-2-27 13:00:33'

然后查询。 请注意，您可以使用任何数据范围，因此，如果要使用一个月的时间，只需将其计算出来并将其放在@startDate和@endDate中

DECLARE @startDate DATETIME='25-Feb-2015'
DECLARE @endDate DATETIME='28-Feb-2015'
DECLARE @numberOfDays INT = DATEDIFF(DAY, @startDate, @endDate)

declare @dayColumns TABLE (delta int, colName varchar(12))

-- Produce 1 row for each day in the report. Note that this is limited by the 
-- number of objects in sysobjects (which is about 2000 so its a high limit)
-- Each row contains a delta date offset, @startDate+delta gives each date to report 
-- which is converted to a valid SQL column name in the format colYYYYMMDD
INSERT INTO @dayColumns (delta, colName)
SELECT delta, 'col'+CONVERT(varchar(12),DATEADD(day,delta,@startDate),112) as colName from (
  select (ROW_NUMBER() OVER (ORDER BY sysobjects.id))-1 as delta FROM sysobjects 
) daysAhead
WHERE delta<=@numberOfDays

-- Create a comma seperated list of columns to report
DECLARE @cols AS NVARCHAR(MAX)= ''
SELECT @cols=CASE WHEN @cols='' THEN @cols ELSE @cols+',' END + colName FROM @dayColumns ORDER BY delta
DECLARE @totalCount AS NVARCHAR(MAX)= ''
SELECT @totalCount=CASE WHEN @totalCount='' THEN '' ELSE @totalCount+' + ' END + 'ISNULL(' + colName +',0)' FROM @dayColumns ORDER BY delta

-- Produce a SQL statement which outputs a variable number of pivoted columns
DECLARE @query AS NVARCHAR(MAX)
SELECT @query=
'declare @days TABLE (reportDay date, colName varchar(12))

INSERT INTO @days (reportDay, colName)
SELECT DATEADD(day,Delta,'''+CONVERT(varchar(22),@startDate,121)+'''), ''col''+CONVERT(varchar(12),DATEADD(day,delta,'''+CONVERT(varchar(22),@startDate,121)+'''),112) as colName from (
  select (ROW_NUMBER() OVER (ORDER BY sysobjects.id))-1 as Delta FROM sysobjects 
) daysAhead
WHERE Delta<='+CAST(@numberOfDays as varchar(10))+'

SELECT pivoted.*,'+@totalCount+' as total FROM (
  SELECT * FROM (
    select data.Name, d.colName, 1 as numRows
    from @days d
    LEFT OUTER JOIN data ON CAST(data.[Date] as DATE)=d.reportDay
  ) as s
  PIVOT (
    SUM(numRows) FOR colName in ('+@cols+')
  ) as pa
) as pivoted
WHERE Name is not null'

-- Run the query
EXEC (@query)

输出为：

    Name                           col20150225 col20150226 col20150227 col20150228 total
    ------------------------------ ----------- ----------- ----------- ----------- -----------
    ITEM A                         2           1           NULL        NULL        3
    ITEM B                         NULL        2           NULL        NULL        2
    ITEM C                         NULL        NULL        1           NULL        1

您可以通过解析演示文稿代码中的列标题（格式为colYYYYMMDD）来确定每列的日期。

Answer 2

您可以在查询中使用PIVOT进行此操作

SELECT name, 
       [1], 
       [2], 
       [3], 
       [4], 
       [5],
       [6],
       [7],
       [8],
       [9],
       [10],
       [11],
       [12],
       [13],
       [14],
       [15],
       [16],
       [17],
       [18],
       [19], 
       [20], 
       [21], 
       [22], 
       [23], 
       [24], 
       [25], 
       [26], 
       [27], 
       [28], 
       [29], 
       [30], 
       [31],
       ([1] + [2] + [3] + [4] + [5] + [6] + [7] + [8] + [9] + [10] + [11] + [12] + [13] + [14] + [15] + [16] + [17] + [18] + [19] +  [20] + [21] + [22] + [23] + [24] + [25] + [26] + [27] + [28] + [29] + [30] + [31]) as total
FROM   
(
    SELECT Name, 
        id, 
        Datepart(day, [date]) day 
    FROM   item 
    WHERE  MONTH([date]) = 2 AND YEAR([date]) = 2015
) x 
PIVOT 
(
    count(id) 
    FOR day IN ([1], [2], [3], [4], [5], [6], [7], [8], [9], [10], [11], [12], [13], [14], [15], [16], [17], [18], [19],  [20], [21], [22], [23], [24], [25], [26], [27], [28], [29], [30], [31]) 
) p 

Answer 3

select convert(varchar,PaymentDate,103) AS date,
       DatePart(MONTH,PaymentDate) as month,
       CAST(SUM(Total_Amount) as INT) as Revenue 
from Tbl_Name 
where YEAR(PaymentDate) = YEAR(CURRENT_TIMESTAMP)
  AND MONTH(PaymentDate) = MONTH(CURRENT_TIMESTAMP) 
GROUP BY convert(varchar,PaymentDate,103),
         DatePart(MONTH,PaymentDate) 
order by date;