我如何将工作时间转换为布尔值？

Question

如何转换工作时间 08:00:00-11:59:00;13:00:00-16:59:00;转换为 48 位布尔格式，如

 "000000000000000011111111001111111100000000000000"

其中每个数字是指使用 Oracle SQL 查询的 30 分钟粒度？

Answer 1

假设您从一个字符串开始，该字符串始终以分号分隔 from/to 对，最后一对后有一个尾随分号；那些时间总是 HH24:MI:SS ，秒数总是零，如图所示；然后...您可以将字符串拆分为多个字符串对，代表每个 from/to 对：

select regexp_substr('08:00:00-11:59:00;13:00:00-16:59:00;', '(.*?)(;|-|$)', 1, (2 * level - 1), null, 1),
  regexp_substr('08:00:00-11:59:00;13:00:00-16:59:00;', '(.*?)(;|-|$)', 1, 2 * level, null, 1)
from dual
connect by level <= regexp_count('08:00:00-11:59:00;13:00:00-16:59:00;', ';')

REGEXP_S REGEXP_S
-------- --------
08:00:00 11:59:00
13:00:00 16:59:00

并且您可以在标称的一天内生成所有半小时块（选择一个不受 DST 转换影响的块）：

select to_char(date '2000-01-01' + ((level - 1) / 48), 'HH24:MI":00"')
from dual
connect by level <= 48

TO_CHAR(
--------
00:00:00
00:30:00
01:00:00
01:30:00
02:00:00
...
23:00:00
23:30:00

然后使用字符串比较（这就是时间格式很重要的原因）将它们连接在一起以查看重叠的地方；为简单起见，使用 CTE 提供初始字符串，然后提供前两个查询的结果：

with t1 (working_hours) as (
  select '08:00:00-11:59:00;13:00:00-16:59:00;' from dual
),
t2 (working_from, working_to) as (
  select regexp_substr(working_hours, '(.*?)(;|-|$)', 1, (2 * level - 1), null, 1),
    regexp_substr(working_hours, '(.*?)(;|-|$)', 1, 2 * level, null, 1)
  from t1
  connect by level <= regexp_count(working_hours, ';')
),
t3 (block_from) as (
  select to_char(date '2000-01-01' + ((level - 1) / 48), 'HH24:MI":00"')
  from dual
  connect by level <= 48
)
select block_from,
  case when t2.working_from is null then 0 else 1 end as flag
from t3
left join t2 on t2.working_from <= t3.block_from and t2.working_to >= t3.block_from

BLOCK_FROM  FLAG
----------  ----
00:00:00    0
00:30:00    0
...
07:30:00    0
08:00:00    1
08:30:00    1
...
11:00:00    1
11:30:00    1
12:00:00    0
12:30:00    0
13:00:00    1
13:30:00    1
...
16:00:00    1
16:30:00    1
17:00:00    0
...
23:00:00    0
23:30:00    0

然后最终将它们聚合成一个结果字符串：

with t1 (working_hours) as (
  select '08:00:00-11:59:00;13:00:00-16:59:00;' from dual
),
t2 (working_from, working_to) as (
  select regexp_substr(working_hours, '(.*?)(;|-|$)', 1, (2 * level - 1), null, 1),
    regexp_substr(working_hours, '(.*?)(;|-|$)', 1, 2 * level, null, 1)
  from t1
  connect by level <= regexp_count(working_hours, ';')
),
t3 (block_from) as (
  select to_char(date '2000-01-01' + ((level - 1) / 48), 'HH24:MI":00"')
  from dual
  connect by level <= 48
)
select listagg(case when t2.working_from is null then 0 else 1 end)
  within group (order by t3.block_from) as result
from t3
left join t2 on t2.working_from <= t3.block_from and t2.working_to >= t3.block_from

RESULT
------------------------------------------------
000000000000000011111111001111111100000000000000

db<>fiddle

如果您的初始字符串实际上来自 table 并且您需要一次对多行进行此转换，那么连接拆分会稍微复杂一些，递归 CTE 可能更合适table 那部分。

纯属娱乐，here's an example。

我如何将工作时间转换为布尔值？

how do i convert the working hours to Boolean?

sql

time

oracle-sqldeveloper

boolean-logic

granularity