Question

plz Solve My Table test

Id int
Emp_ID int
ontime Datetime

从测试中选择*

Id  Emp_ID  ontime 
1   1   2013-02-11 2:14:25.387  
2   2   2013-02-11 14:14:25.387
1   1   2013-02-11 20:14:25.387
3   3   2013-02-13 10:14:25.387
4   2   2013-02-13 10:14:25.387
5   1   2013-02-13 10:14:25.387
6   1   2013-02-14 11:14:25.387

我的要求这个

用户输入两个日期并记录每日显示计数

只记录

日期1 = 11/2/2013 日期2 = 13/2/2013

Emp_ID  11/2/2013   12/2/2013   13/2/2013
1       2       0       1
2       1       0       1
3       0       0       1

thx

THX

但是这一个tbl_emp

Emp_ID int Emp_name varchar

| EMP_ID | 2013-02-11 | 2013-02-12 | 2013-02-13 |
-------------------------------------------------
|      1 |          2 |          0 |          1 |
|      2 |          1 |          0 |          1 |
|      3 |          0 |          0 |          1 |
|      4 |          0 |          0 |          0 |

Emp_ID 4未插入测试表

Answer 1

此类数据转换称为PIVOT。从SQL Server 2005开始，有一个函数可以为您执行此操作。

如果您提前知道这些值，那么您可以对查询进行硬编码，类似于：

select *
from
(
  select id,
    [Emp_ID], 
    cast([ontime] as date) ontime
  from test
) src
pivot
(
  count(id)
  for ontime in ([2013-02-11], [2013-02-12], [2013-02-13])
) piv

请参阅SQL Fiddle with Demo。

如果要转换的列数未知，则需要使用动态SQL。这可以通过使用WHERE子句进一步过滤来仅查找某些日期：

DECLARE @cols AS NVARCHAR(MAX),
    @query  AS NVARCHAR(MAX)

select @cols = STUFF((SELECT distinct ',' + QUOTENAME(cast([ontime] as date)) 
                    from test
            FOR XML PATH(''), TYPE
            ).value('.', 'NVARCHAR(MAX)') 
        ,1,1,'')

set @query = 'SELECT emp_id,' + @cols + ' from 
             (
                select Emp_ID pivId,
                  [Emp_ID], 
                  cast([ontime] as date) ontime
                from test
            ) x
            pivot 
            (
                count(pivId)
                for ontime in (' + @cols + ')
            ) p '

execute(@query)

请参阅SQL Fiddle with Demo。

如果您想保证您将返回所需的日期，则可以先生成日期列表。因此，整个动态SQL脚本将是：

declare @startdate datetime
declare @enddate datetime

set @startdate = '2013-02-11'
set @enddate = '2013-02-13'

;with cte (startdate) as
(
  select @startdate
  union all
  select DATEADD(dd, 1, startdate)
  from cte
  where DATEADD(dd, 1, startdate) <= @enddate
)
select *
into #tempDates
from cte;    

DECLARE @cols AS NVARCHAR(MAX),
    @query  AS NVARCHAR(MAX)

select @cols = STUFF((SELECT distinct ',' + QUOTENAME(cast(startdate as date)) 
                    from #tempDates
            FOR XML PATH(''), TYPE
            ).value('.', 'NVARCHAR(MAX)') 
        ,1,1,'')

set @query = 'SELECT emp_id,' + @cols + ' from 
             (
                select Emp_ID pivId,
                  [Emp_ID], 
                  cast([ontime] as date) ontime
                from test
            ) x
            pivot 
            (
                count(pivId)
                for ontime in (' + @cols + ')
            ) p '

execute(@query)

见SQL Fiddle with Demo。这将返回结果：

| EMP_ID | 2013-02-11 | 2013-02-12 | 2013-02-13 |
-------------------------------------------------
|      1 |          2 |          0 |          1 |
|      2 |          1 |          0 |          1 |
|      3 |          0 |          0 |          1 |

编辑＃1，如果你想包括所有员工，那么你会希望在另一个表上使用JOIN，你的代码将类似于：

declare @startdate datetime
declare @enddate datetime

set @startdate = '2013-02-11'
set @enddate = '2013-02-13'

;with cte (startdate) as
(
  select @startdate
  union all
  select DATEADD(dd, 1, startdate)
  from cte
  where DATEADD(dd, 1, startdate) <= @enddate
)
select *
into #tempDates
from cte;    

DECLARE @cols AS NVARCHAR(MAX),
    @query  AS NVARCHAR(MAX)

select @cols = STUFF((SELECT distinct ',' + QUOTENAME(cast(startdate as date)) 
                    from #tempDates
            FOR XML PATH(''), TYPE
            ).value('.', 'NVARCHAR(MAX)') 
        ,1,1,'')

set @query = 'SELECT emp_id, Emp_name,' + @cols + ' from 
             (
                select e.Emp_ID pivId,
                  e.[Emp_ID], 
                  e.Emp_name,
                  cast([ontime] as date) ontime
                from Emp_Table e
                left join test t
                  on e.emp_id = t.emp_id
            ) x
            pivot 
            (
                count(pivId)
                for ontime in (' + @cols + ')
            ) p '

execute(@query)

请参阅SQL Fiddle with Demo

输入两个日期和记录显示每日计数明智

1 个答案: