Question

我有两张桌子，一张是工作模式，另一张是缺席。

1）工作模式

ID | Shift Start | Shift End
123| 01-03-2017  | 02-03-2017

2）缺席

ID| Absence Start | Absence End
123| 01-03-2017   | 04-03-2017

从工作模式中选择行时，排除任何在缺席表中标记为缺席的日期的最佳方法是什么？

例如，我有一份报告，它使用工作模式表来计算员工每周工作的天数，但我不希望它包含已标记为缺勤的日期。缺席表是否有意义？也不希望它包括在缺席开始和缺席结束日期之间的任何日期？

Answer 1

如果{ ContentType: "image/jpeg; charset=utf-8", 'x-amz-acl': 'public-read' }的范围始终包含要排除的班次，您可以使用absence：

not exists()

rextester演示：http://rextester.com/DCODC76816

返回：

select *
from WorkPatterns w
where not exists (
  select 1
  from Absences a
  where a.Id = w.Id
    and a.AbsenceStart <= w.ShiftStart
    and a.AbsenceEnd   >= w.ShiftEnd
  )

给出此测试设置：

+-----+------------+------------+
| id  | ShiftStart |  ShiftEnd  |
+-----+------------+------------+
| 123 | 2017-02-27 | 2017-02-28 |
| 123 | 2017-03-05 | 2017-03-06 |
+-----+------------+------------+

Answer 2

从工作模式中选择行时，最好的方法是什么

如果您只处理白色日期（没有时间）并控制数据库架构，一种方法是创建日历表，您将在公司成立以来的所有日期和未来几年的所有日期填写该表一次。之后很容易加入其他表格的白日期并做数学。

如果您在构建TSQL查询时遇到问题，请编辑问题，详细了解表格，关系和所需结果的列和值。

Answer 3

这个怎么样：

SELECT WP_START.[id], WP_START.[shift_start], WP_START.[shift_end]
FROM work_pattern AS WP_START
INNER JOIN absences AS A ON WP_START.id = A.id 
WHERE WP_START.[shift_start] NOT BETWEEN A.[absence_start] AND A.[absence_end]

UNION

SELECT WP_END.[id], WP_END.[shift_start], WP_END.[shift_end]
FROM work_pattern AS WP_END
INNER JOIN absences AS A ON WP_END.id = A.id 
WHERE WP_END.[shift_end] NOT BETWEEN A.[absence_start] AND A.[absence_end]

在SQL小提琴上看到它：http://sqlfiddle.com/#!6/49ae6/6

Answer 4

这是我的示例，其中包含Date Dimension表。如果您的DBA不能添加它，您可以创建#dateDim作为临时表，就像我用SQLFiddle 做的那样（我不知道我能做什么那）。典型的日期维度将包含有关日期的更多详细信息，但如果无法添加表格，只需使用您需要的内容即可。您必须填写您需要的其他假期。我经常使用的DateDim是https://github.com/shawnoden/SQL_Stuff/blob/master/sql_CreateDateDimension.sql

SQL Fiddle

MS SQL Server 2014架构设置：

/* Tables for your test data. */
CREATE TABLE WorkPatterns ( id int, ShiftStart date, ShiftEnd date ) ;
INSERT INTO WorkPatterns ( id, ShiftStart, ShiftEnd )
VALUES
      (123, '20170101', '20171031')
    , (124, '20170601', '20170831')
;

CREATE TABLE Absences ( id int, AbsenceStart date, AbsenceEnd date ) ;
INSERT INTO Absences ( id, AbsenceStart, AbsenceEnd )
VALUES 
      ( 123, '20170123', '20170127' )
    , ( 123, '20170710', '20170831' )
    , ( 124, '20170801', '20170820' )
;

/* ******** MAKE SIMPLE CALENDAR TABLE ******** */

CREATE TABLE dateDim (
      theDate             DATE        NOT NULL
    , IsWeekend           BIT         DEFAULT 0
    , IsHoliday           BIT         DEFAULT 0
    , IsWorkDay           BIT         DEFAULT 0
);

/* Populate basic details of dates. */
INSERT dateDim(theDate, IsWeekend, IsHoliday) 
SELECT d
  , CONVERT(BIT, CASE WHEN DATEPART(dw,d) IN (1,7) THEN 1 ELSE 0 END)
  , CONVERT(BIT, CASE WHEN d = '20170704' THEN 1 ELSE 0 END) /* 4th of July. */
FROM (
    SELECT d = DATEADD(DAY, rn - 1, '20170101')
    FROM 
    (
        SELECT TOP (DATEDIFF(DAY, '20170101', '20171231')) 
            rn = ROW_NUMBER() OVER (ORDER BY s1.[object_id])
        FROM sys.all_objects AS s1
        CROSS JOIN sys.all_objects AS s2
        ORDER BY s1.[object_id]
    ) AS x
) AS y ;

/* If not a weekend or holiday, it's a WorkDay. */
UPDATE dateDim 
SET IsWorkDay = CASE WHEN IsWeekend = 0 AND IsHoliday = 0 THEN 1 ELSE 0 END
;

查询计算：

SELECT wp.ID, COUNT(d.theDate) AS workDayCount
FROM WorkPatterns wp
INNER JOIN dateDim d ON d.theDate BETWEEN wp.ShiftStart AND wp.ShiftEnd
  AND d.IsWorkDay = 1
LEFT OUTER JOIN Absences a ON d.theDate BETWEEN a.AbsenceStart AND a.AbsenceEnd
  AND wp.ID = a.ID
WHERE a.ID IS NULL
GROUP BY wp.ID
ORDER BY wp.ID

<强> Results ：

|  ID | workDayCount |
|-----|--------------|
| 123 |          172 |  << 216 total days, 44 non-working
| 124 |           51 |  <<  65 total days, 14 non-working

排除另一个表

4 个答案: