以当天为搜索点如何获取同月前几周的数据

Question

如何知道该月的前一周。

例如，如果我 table 是这样的：

emp_num   trans_date 
  22       1-10-2015
  22       5-10-2015
  22       7-10-2015
  22       11-10-2015
  22       14-10-2015
  22       19-10-2015
  22       27-10-2015

现在每个月有 4 周。

所以我有一个像 11-10-2015 这样的约会（在第二周）我想获得同月所有前几周的结果。所以在这个例子中我想要十月第一周的结果是这样的：

emp_num   trans_date 
  22       1-10-2015
  22       5-10-2015
  22       7-10-2015

注：

The start day of the week is: sat

The end day of the week is:Fri

Answer 1

我没有在 informix 工作过，但在 SQL 中是这样的：

DECLARE @SomeDate DATE
SET @SomeDate = '2015-10-11'  

DECLARE @DayOfWeek INT
SELECT @DayOfWeek = ((DATEPART(dw, @SomeDate) ) % 7)

DECLARE @WeekStart DATE
SELECT @WeekStart = DATEADD(d, -@DayOfWeek, @SomeDate)

SELECT * FROM SomeTable 
WHERE MONTH(trans_date) = MONTH(@SomeDate) AND trans_date < @WeekStart

Answer 2

如果我对你的问题的理解正确，对于 2015 年 10 月，你预计会出现以下结果：

RunDate     Weekday  Result
01-10-2015  Thurs    [nothing]
02-10-2015  Fri      [nothing]
03-10-2015  Sat      01-10-2015, 02-10-2015
04-10-2015  Sun      01-10-2015, 02-10-2015 (same as above)
...
09-10-2015  Fri      01-10-2015, 02-10-2015 (same as above)
10-10-2015  Sat      01-10-2015, ... , 09-10-2015
...
16-10-2015  Fri      01-10-2015, ... , 09-10-2015 (same as above)
17-10-2015  Sat      01-10-2015, ... , 16-10-2015
20-10-2015  Tue      01-10-2015, ... , 16-10-2015 (same as above)
30-10-2015  Fri      01-10-2015, ... , 23-10-2015
31-10-2015  Sat      01-10-2015, ... , 30-10-2015

如果是这种情况，那么以下查询将执行您的要求：

SELECT trans_date, emp_num, COUNT(1)
  FROM transaction_table
  WHERE MONTH(trans_date) = MONTH(TODAY)
    AND YEAR(trans_date) = YEAR(TODAY)
    AND trans_date <= DECODE(WEEKDAY(TODAY),
                                6, TODAY -1,
                                TODAY - WEEKDAY(TODAY) -2)
GROUP BY 1, 2
ORDER BY 1, 2

DECODE() 中的逻辑是管理与 WEEKDAY() 默认行为的偏移量，其中 Sun => 0 和 Sat => 6。关键字 TODAY 在上面的查询可以替换为变量或任意单个日期。

以当天为搜索点如何获取同月前几周的数据

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