Question

我有一个选择查询，它选择了以下两个日期/时间段之间的数据

SELECT *
FROM table1
WHERE table1 .row_creation > TO_DATE ('2017-04-03 11:00:00 AM','yyyy-mm-dd hh:mi:ss pm')
AND table1 .row_creation   < TO_DATE('2017-04-04 03:00:00 AM','yyyy-mm-dd hh:mi:ss pm')
AND NOT EXISTS 
  (SELECT *
  FROM table1
  WHERE                    <cond>
  AND table1 .row_creation > TO_DATE ('2017-04-03 11:00:00 AM','yyyy-mm-dd hh:mi:ss pm')
  AND table1 .row_creation < TO_DATE('2017-04-04 03:00:00 AM','yyyy-mm-dd hh:mi:ss pm')
  )

我想修改此查询以一次获取一个月中所有连续工作日的数据。谁能给我一些指导？

Answer 1

如果愿意，可以在group by子句中添加更多列。

SELECT TO_CHAR(row_creation,'W') FROM table1 WHERE table1 .row_creation between TO_DATE ('2017-04-03 11:00:00 AM','yyyy-mm-dd hh:mi:ss pm') AND TO_DATE('2017-04-04 03:00:00 AM','yyyy-mm-dd hh:mi:ss pm') AND NOT EXISTS ( SELECT * FROM table1 WHERE AND table1.row_creation BETWEEN TO_DATE ('2017-04-03 11:00:00 AM','yyyy-mm- dd hh:mi:ss pm') AND TO_DATE('2017-04-04 03:00:00 AM','yyyy-mm-dd hh:mi:ss pm') ) AND TO_CHAR(row_creation,'DAY') NOT IN ('SATURDAY','SUNDAY') GROUP BY TO_CHAR(row_creation,'W')

Answer 2

我认为您可以简化整个2017年4月的操作，例如：

select * 
  from (
    select a.*, 
           count(case when cond='X' then 1 end) over (partition by trunc(row_creation-3/24)) cnt
      from table1 a
      where row_creation - 3/24 between date '2017-04-01' and date '2017-05-01'
        and to_char(row_creation, 'hh24') not between '03' and '10')
  where cnt = 0

其中

row_creation - 3/24

将日期设置为00:00-03:00至前一天，

count(case when <cond> then 1 end) over (partition by trunc(row_creation - 3/24))

计算当天的不需要的行，最后：

where cnt = 0

如果当天至少有一个不需要的案件，

负责消除行。

^{dbfiddle example}

概括查询以获取给定范围内的所有数据

2 个答案: