如何排除 SQL 中的周末？

Question

我在 Orders table 中有一个日期列 SLA_Date。

我的 SLA_Date 应该不包括周末（周六和周日）。应单独显示工作日的数据。

如何在 SQL 中执行此操作？

Answer 1

您可以尝试以下条件来过滤您的数据：

WHERE case to_char (SLA_Date, 'FmDay', 'nls_date_language=english') <> 'Saturday'
  AND case to_char (SLA_Date, 'FmDay', 'nls_date_language=english') <> 'Sunday'

Answer 2

select *
from   orders
where  TRUNC (thing) - TRUNC (thing, 'IW') between 0 and 4;

http://sqlfiddle.com/#!4/db891/1

Answer 3

您只需添加以下过滤器:

WHERE TO_CHAR(date_column, 'DY','NLS_DATE_LANGUAGE=AMERICAN') NOT IN ('SAT', 'SUN')

您的查询如下：

SELECT SLA_Date
   FROM orders
 WHERE TO_CHAR(SLA_Date, 'DY','NLS_DATE_LANGUAGE=AMERICAN') NOT IN ('SAT', 'SUN')

例如（WITH子句只是为了构建一个测试用例），下面的查询是只显示从1st May 2015到31st May 2015的工作日（即不包括周六和周日） :

SQL> WITH DATA AS
  2    (SELECT to_date('05/01/2015', 'MM/DD/YYYY') date1,
  3      to_date('05/31/2015', 'MM/DD/YYYY') date2
  4    FROM dual
  5    )
  6  SELECT date1+LEVEL-1 the_date,
  7        TO_CHAR(date1+LEVEL-1, 'DY','NLS_DATE_LANGUAGE=AMERICAN') day
  8  FROM DATA
  9  WHERE TO_CHAR(date1+LEVEL-1, 'DY','NLS_DATE_LANGUAGE=AMERICAN')
 10        NOT IN ('SAT', 'SUN')
 11  CONNECT BY LEVEL <= date2-date1+1;

THE_DATE  DAY
--------- ---
01-MAY-15 FRI
04-MAY-15 MON
05-MAY-15 TUE
06-MAY-15 WED
07-MAY-15 THU
08-MAY-15 FRI
11-MAY-15 MON
12-MAY-15 TUE
13-MAY-15 WED
14-MAY-15 THU
15-MAY-15 FRI
18-MAY-15 MON
19-MAY-15 TUE
20-MAY-15 WED
21-MAY-15 THU
22-MAY-15 FRI
25-MAY-15 MON
26-MAY-15 TUE
27-MAY-15 WED
28-MAY-15 THU
29-MAY-15 FRI

21 rows selected.

SQL>

有关详细示例，请参阅 Generate Date, Month Name, Week Number, Day number between two dates

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