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社区首页 >问答首页 >如何将经常性工作日创建为枢轴中的列?

问如何将经常性工作日创建为枢轴中的列?
EN

Database Administration用户
提问于 2018-10-22 23:42:58
回答 2查看 1.2K关注 0票数 8

我是编程和数据库方面的新手,希望能在下面的场景中提供一些帮助。

我在SQL Server中使用PHP。我正在构建一个员工出勤系统,我希望创建一个(枢轴)表,其中月份作为行,所有工作日的名称都作为列(用于特定年份)。单元格中的值为日数(1、2、3.31)。

单元格的背景色(已作为表列存在)声明员工休假的类型。该表有以下列:employee_id, leave_date, leave_type, leave_type_color。

我想取得以下结果:

谢谢。

EN

回答 2

Database Administration用户

回答已采纳

发布于 2018-10-23 02:44:25

其中最复杂的部分就是以这种格式构建日历。旋转并用HTML包围它是非常容易的。首先,让我们从这里开始,您的员工表中有休假日期。leave_type似乎与眼前的问题无关。

代码语言:javascript
复制
CREATE TABLE dbo.EmpLeave
(
  EmployeeID int,
  leave_date date,
  leave_type_color char(6)
);

INSERT dbo.EmpLeave(EmployeeID,leave_date,leave_type_color)
  VALUES(1,'2018-01-02','7777cc'),(1,'2018-04-01','ffffac');

我想出的过程如下(以及警告:它假定为@@DATEFIRST = 7):

代码语言:javascript
复制
CREATE PROCEDURE dbo.BuildLeaveHTMLTable
  @EmployeeID int,
  @Year smallint = NULL
AS
BEGIN
  SET NOCOUNT ON;
  SET @Year = COALESCE(@Year, DATEPART(YEAR, GETDATE()));
  DECLARE @FirstDay date = DATEADD(YEAR, @Year-1900, 0);

  ;WITH Numbers AS ( -- 366 possible days (leap year)
    SELECT n = 1 UNION ALL SELECT n + 1 FROM Numbers WHERE n <= 365
  ),
  Calendar AS ( -- a year's worth of dates and dateparts 
    SELECT [Date] = d,
      MonthStart = DATEADD(DAY, 1-DAY(d),d),
      Y  = CONVERT(smallint, DATEPART(YEAR,   d)),
      M  = CONVERT(tinyint,  DATEPART(MONTH,  d)),
      D  = CONVERT(tinyint,  DATEPART(DAY,    d)),
      WY = CONVERT(tinyint,  DATEPART(WEEK,   d)),
      DW = CONVERT(tinyint,  DATEPART(WEEKDAY,d))
    FROM
    (
      SELECT d = CONVERT(date,DATEADD(DAY, n-1, @FirstDay)) FROM Numbers
    ) AS c WHERE YEAR(d) = @Year -- in case it's not a leap year
  ),
  BaseSlots AS ( -- base set of 37 ints 
   -- month can be spread across 6 weeks, but no more than 2 days in 6th week
    SELECT TOP (37) slot = n FROM Numbers ORDER BY n
  ),
  Months AS ( -- base set of 12 ints
    SELECT TOP (12) m = slot FROM BaseSlots ORDER BY slot
  ),
  SlotAlignment AS ( -- align days of week to slot numbers
    -- this is the most cryptic part of this solution
    -- determines which set of 7 slots, and which slot 
    -- exactly, a given date will appear under
    SELECT c.*, slot = DW+(c.WY+1-DATEPART(WEEK,c.MonthStart)-1)*7
      FROM Calendar AS c 
      INNER JOIN Months AS m ON c.M = m.m
  ),
  SlotMatrix AS ( -- extrapolate actual dates to 37 x 12 matrix
    SELECT m.m, s.slot, sa.[Date] 
      FROM BaseSlots AS s 
      CROSS JOIN Months AS m
      LEFT OUTER JOIN SlotAlignment AS sa
      ON sa.m = m.m AND sa.slot = s.slot
  ),
  FinalHTML AS ( -- build some HTML!
    SELECT m = '<!-- ' + RIGHT('0' + RTRIM(m), 2) + ' -->', 
      slot, cell = CASE WHEN slot = 1 THEN '<tr><th>' 
        + COALESCE(DATENAME(MONTH,DATEADD(MONTH, m-1, 0)),'') 
        + '</th>' ELSE '' END + '<td' + COALESCE(' bgcolor=#' 
        + RIGHT(CONVERT(varchar(10),CONVERT(varbinary(8), el.leave_type_color),1),6),
          CASE WHEN DATEPART(WEEKDAY, [Date]) IN (1,7) 
          THEN ' bgcolor=#cccccc' ELSE '' END)
        + '>' + COALESCE(RTRIM(DATEPART(DAY,[Date])), ' ')
        + '</td>' + CASE WHEN slot = 37 THEN '</tr>' ELSE '' END
      FROM SlotMatrix AS q LEFT OUTER JOIN dbo.EmpLeave AS el
      ON q.Date = el.leave_date
      AND el.EmployeeID = @EmployeeID
  ) -- now turn it sideways
  SELECT m = '<!-- 00 -->', 
    [1]  = '<tr><th>Month</th><th>S</th>',    [2]  = '<th>M</th>', 
    [3]  = '<th>T</th>', [4]  = '<th>W</th>', [5]  = '<th>T</th>', 
    [6]  = '<th>F</th>', [7]  = '<th>S</th>', [8]  = '<th>S</th>', 
    [9]  = '<th>M</th>', [10] = '<th>T</th>', [11] = '<th>W</th>',
    [12] = '<th>T</th>', [13] = '<th>F</th>', [14] = '<th>S</th>', 
    [15] = '<th>S</th>', [16] = '<th>M</th>', [17] = '<th>T</th>',
    [18] = '<th>W</th>', [19] = '<th>T</th>', [20] = '<th>F</th>', 
    [21] = '<th>S</th>', [22] = '<th>S</th>', [23] = '<th>M</th>',
    [24] = '<th>T</th>', [25] = '<th>W</th>', [26] = '<th>T</th>', 
    [27] = '<th>F</th>', [28] = '<th>S</th>', [29] = '<th>S</th>', 
    [30] = '<th>M</th>', [31] = '<th>T</th>', [32] = '<th>W</th>', 
    [33] = '<th>T</th>', [34] = '<th>F</th>', [35] = '<th>S</th>',
    [36] = '<th>S</th>', [37] = '<th>M</th>'
  UNION ALL
  (
    SELECT * FROM FinalHTML PIVOT (MAX(cell) FOR slot 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],[32],[33],[34],[35],[36],[37]
    )) AS p
  )
  ORDER BY m OPTION (MAXRECURSION 366);
END
GO

这一呼吁的结果:

代码语言:javascript
复制
EXEC dbo.BuildLeaveHTMLTable @EmployeeID = 1;

如下所示(我在第7天专栏文章中停了下来):

您必须自己添加<table> / </table>包装器,但下面是将输出放入其中并保存为HTML时的样子(当然,您还可以使用CSS进一步增强它):

当假期落在一个周末,离开的颜色超过周末的颜色,但这是很容易调整的。改变这一点:

代码语言:javascript
复制
  + COALESCE(' bgcolor=#' + RTRIM(el.leave_type_color),
      CASE WHEN DATEPART(WEEKDAY, [Date]) IN (1,7) 
      THEN ' bgcolor=#cccccc' ELSE '' END)

对此:

代码语言:javascript
复制
  + CASE WHEN DATEPART(WEEKDAY, [Date]) IN (1,7) 
      THEN ' bgcolor=#cccccc' ELSE COALESCE(' bgcolor=#' 
      + RTRIM(el.leave_type_color), '') END

要将十进制格式的颜色(如65280)转换为其RGB等效(00FF00),您必须进行大量的操作。首先,我会考虑将其存储为RGB十六进制,但我在这里更新了解决方案,类似于以下内容:

代码语言:javascript
复制
SELECT RIGHT(CONVERT(varchar(10),CONVERT(varbinary(8), 65280),1),6);
票数 11
EN

Database Administration用户

发布于 2018-10-23 00:28:13

首先考虑你想要的列,这基本上是“第一周第一天(太阳)”,“第一天第二周(星期一)”,直到“第六天第七周(星期六)”。基本上,第一天到第四十二天。1月1日是一月的“第一周第二天”。我现在叫它WeekPlusDay。

要想弄清楚每一天从哪里开始,只需考虑工作日的部分时间。

然后,您的数据集只需包含“WeekPlusDay”值,就可以显示DayOfMonth。

票数 1
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页面原文内容由Database Administration提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接:

https://dba.stackexchange.com/questions/220772

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