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。
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。
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):
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:
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:
+ COALESCE(' bgcolor=#' + RTRIM(el.leave_type_color),
CASE WHEN DATEPART(WEEKDAY, [Date]) IN (1,7)
THEN ' bgcolor=#cccccc' ELSE '' END)对此:
+ 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:
SELECT RIGHT(CONVERT(varchar(10),CONVERT(varbinary(8), 65280),1),6);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。
https://dba.stackexchange.com/questions/220772
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